Quadratic Relations and Equations

Quadratic Factoring Techniques

Factoring is the process of rewriting a polynomial as a product of simpler expressions. For quadratics, factoring is especially useful because it helps us solve equations, find x-intercepts, and better understand the structure of a parabola.

A quadratic expression usually has the form \(ax^2+bx+c\), where \(a\neq 0\). In this lesson, you will learn how to factor several common types of quadratics:

• Quadratic trinomials with leading coefficient 1
• Quadratic trinomials with leading coefficient greater than 1
• Perfect square trinomials
• Differences of squares

Before trying any factoring method, it is always smart to look for a greatest common factor (GCF). Factoring out the GCF first makes the remaining expression easier to work with.

For example, in \(6x^2+12x\), both terms share a factor of \(6x\):

$$6x^2+12x=6x(x+2)$$

Once the GCF is removed, you can focus on the remaining quadratic expression.

1. Factoring trinomials when the leading coefficient is 1

These are quadratics of the form:

$$x^2+bx+c$$

To factor this kind of trinomial, find two numbers that:

• multiply to give \(c\)
• add to give \(b\)

Then write the factorization as:

$$x^2+bx+c=(x+m)(x+n)$$

where \(m\) and \(n\) are the two numbers you found.

Worked Example 1

Factor:

$$x^2+7x+12$$

We need two numbers that multiply to \(12\) and add to \(7\).

• \(1\) and \(12\) add to \(13\)
• \(2\) and \(6\) add to \(8\)
• \(3\) and \(4\) add to \(7\)

So the correct pair is \(3\) and \(4\).

$$x^2+7x+12=(x+3)(x+4)$$

You can check by expanding:

$$ (x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12 $$

Sign patterns to remember

• If \(c\) is positive and \(b\) is positive, both signs are positive.
• If \(c\) is positive and \(b\) is negative, both signs are negative.
• If \(c\) is negative, the signs are different.

For example:

• \(x^2-9x+20=(x-5)(x-4)\)
• \(x^2+x-12=(x+4)(x-3)\)

2. Factoring trinomials when the leading coefficient is greater than 1

Now consider expressions of the form:

$$ax^2+bx+c$$

when \(a>1\). These are a little harder because the first term has more than one possible factor pair.

A reliable method is called the product-sum method or grouping method.

Steps:

1. Multiply \(a\cdot c\).
2. Find two numbers that multiply to \(ac\) and add to \(b\).
3. Rewrite the middle term \(bx\) using those two numbers.
4. Factor by grouping.

Worked Example 2

Factor:

$$2x^2+7x+3$$

Step 1: Multiply \(a\cdot c\):

$$2\cdot 3=6$$

Step 2: Find two numbers that multiply to \(6\) and add to \(7\).

Those numbers are \(6\) and \(1\).

Step 3: Rewrite the middle term:

$$2x^2+7x+3=2x^2+6x+x+3$$

Step 4: Factor by grouping:

$$2x^2+6x+x+3=2x(x+3)+1(x+3)$$

Now factor out the common binomial:

$$2x(x+3)+(x+3)=(2x+1)(x+3)$$

So the factorization is:

$$2x^2+7x+3=(2x+1)(x+3)$$

Worked Example 3

Factor:

$$3x^2-x-2$$

Step 1: Multiply \(a\cdot c\):

$$3\cdot (-2)=-6$$

Step 2: Find two numbers that multiply to \(-6\) and add to \(-1\).

Those numbers are \(-3\) and \(2\).

Step 3: Rewrite the middle term:

$$3x^2-x-2=3x^2-3x+2x-2$$

Step 4: Factor by grouping:

$$3x(x-1)+2(x-1)$$

Factor out \((x-1)\):

$$3x(x-1)+2(x-1)=(3x+2)(x-1)$$

So:

$$3x^2-x-2=(3x+2)(x-1)$$

3. Perfect square trinomials

Some quadratics follow special patterns. A perfect square trinomial looks like one of these:

$$a^2+2ab+b^2=(a+b)^2$$ $$a^2-2ab+b^2=(a-b)^2$$

To recognize a perfect square trinomial, check for these three things:

• The first term is a perfect square.
• The last term is a perfect square.
• The middle term is twice the product of the square roots of the first and last terms.

Examples:

• \(x^2+10x+25=(x+5)^2\) because \(25=5^2\) and \(10x=2(x)(5)\)
• \(x^2-14x+49=(x-7)^2\) because \(49=7^2\) and \(-14x=-2(x)(7)\)

Be careful: not every trinomial with square terms is a perfect square trinomial. Always check the middle term.

4. Difference of squares

A difference of squares has the form:

$$a^2-b^2$$

It factors as:

$$a^2-b^2=(a-b)(a+b)$$

This pattern only works when there are two terms with a subtraction sign between them, and both terms are perfect squares.

Worked Example 4

Factor:

$$9x^2-16$$

Both terms are perfect squares:

• \(9x^2=(3x)^2\)
• \(16=4^2\)

So this is a difference of squares:

$$9x^2-16=(3x-4)(3x+4)$$

Another example:

$$x^2-49=(x-7)(x+7)$$

Important note: a sum of squares, such as \(x^2+16\), does not factor using this pattern.

5. A strategy for deciding how to factor

When you see a quadratic expression, use this order:

1. Factor out the GCF if there is one.
2. Check whether it is a difference of squares.
3. Check whether it is a perfect square trinomial.
4. If it is a trinomial, use the appropriate trinomial factoring method.

For example, factor:

$$4x^2-25$$

This has two terms, and both are perfect squares:

$$4x^2=(2x)^2, \quad 25=5^2$$

So:

$$4x^2-25=(2x-5)(2x+5)$$

Now factor:

$$x^2+6x+9$$

The first and last terms are perfect squares, and the middle term is twice the product of their roots:

$$x^2+6x+9=(x+3)^2$$

6. Checking your factored answer

A good habit is to multiply your factors back together. If the product matches the original expression, your factorization is correct.

For example, check:

$$ (3x+2)(x-1) $$

Expand:

$$3x^2-3x+2x-2=3x^2-x-2$$

This matches the original quadratic, so the factorization is correct.

7. Common mistakes to avoid

• Forgetting the GCF: For example, \(2x^2+8x=2x(x+4)\), not just \((x+4)\).
• Using difference of squares on a sum: \(x^2+9\) does not factor as \((x+3)(x+3)\).
• Ignoring signs: Always check whether the numbers should be positive, negative, or one of each.
• Not checking your answer: Expanding the factors can catch errors quickly.

Brief Summary

Factoring quadratics means rewriting them as a product of simpler expressions. Start by factoring out any GCF. Then look for special patterns like perfect square trinomials and differences of squares. If the expression is a trinomial, use number pairs when \(a=1\), or use the product-sum and grouping method when \(a>1\). With practice, you will get faster at recognizing which method to use.

Completing the Square is an algebra method used to rewrite a quadratic expression so it becomes easier to understand and solve. It helps turn a quadratic from standard form, such as \(ax^2+bx+c\), into a form that shows a perfect square.

This is especially useful when solving quadratic equations, finding the vertex of a parabola, and rewriting a quadratic in vertex form.

In this lesson, you will learn what completing the square means, why it works, and how to do it step by step.

1. Review: Forms of a Quadratic

A quadratic can be written in different forms.

• Standard form: \(y=ax^2+bx+c\)
• Vertex form: \(y=a(x-h)^2+k\)

Vertex form is useful because you can quickly identify the vertex of the parabola as \((h,k)\).

Completing the square is the process of changing a quadratic from standard form into a form involving a squared binomial.

2. The Key Idea Behind Completing the Square

A perfect square trinomial has the form

$$x^2+2ax+a^2=(x+a)^2$$

For example:

• \(x^2+6x+9=(x+3)^2\)
• \(x^2-8x+16=(x-4)^2\)

Notice the pattern:

1. Take the coefficient of \(x\).
2. Divide it by 2.
3. Square the result.

That number is the one needed to make a perfect square trinomial.

For example, in \(x^2+10x\):

$$\left(\frac{10}{2}\right)^2=5^2=25$$

So,

$$x^2+10x+25=(x+5)^2$$

3. Completing the Square When the Coefficient of \(x^2\) Is 1

Suppose we want to rewrite

$$x^2+6x+2$$

First, focus on the \(x^2+6x\) part.

Take half of 6 and square it:

$$\left(\frac{6}{2}\right)^2=3^2=9$$

Add and subtract 9 so the value of the expression does not change:

$$x^2+6x+2=x^2+6x+9-9+2$$

Now group the perfect square trinomial:

$$=(x+3)^2-7$$

So the quadratic in vertex form is

$$y=(x+3)^2-7$$

This tells us the vertex is \((-3,-7)\).

4. Why We Add and Subtract the Same Number

When changing an expression, we must keep its value the same. If we add 9, we must also subtract 9.

That way, the total change is 0:

$$+9-9=0$$

This allows us to create a perfect square without changing the original quadratic.

5. Completing the Square to Solve an Equation

Completing the square is also a method for solving quadratic equations.

Suppose we want to solve

$$x^2+8x+5=0$$

First, move the constant term to the other side:

$$x^2+8x=-5$$

Now complete the square. Half of 8 is 4, and \(4^2=16\). Add 16 to both sides:

$$x^2+8x+16=-5+16$$ $$ (x+4)^2=11 $$

Take the square root of both sides:

$$x+4=\pm\sqrt{11}$$

So,

$$x=-4\pm\sqrt{11}$$

This gives the two solutions:

$$x=-4+\sqrt{11} \quad \text{and} \quad x=-4-\sqrt{11}$$

6. Completing the Square When the Coefficient of \(x^2\) Is Not 1

If the coefficient of \(x^2\) is not 1, first factor that coefficient out of the \(x^2\) and \(x\) terms.

For example, rewrite

$$y=2x^2+12x+5$$

Factor out 2 from the first two terms:

$$y=2(x^2+6x)+5$$

Now complete the square inside the brackets. Half of 6 is 3, and \(3^2=9\).

Add and subtract 9 inside the brackets:

$$y=2(x^2+6x+9-9)+5$$

Group the perfect square:

$$y=2((x+3)^2-9)+5$$

Now simplify:

$$y=2(x+3)^2-18+5$$ $$y=2(x+3)^2-13$$

This is vertex form, and the vertex is \((-3,-13)\).

Important: When a number is factored outside the brackets, any number added inside the brackets is also affected by that factor. That is why careful simplification matters.

7. Step-by-Step Method

Here is a general method for completing the square.

To rewrite a quadratic expression in vertex form:

1. If needed, factor out the coefficient of \(x^2\) from the first two terms.
2. Take half of the coefficient of \(x\).
3. Square that number.
4. Add and subtract that value.
5. Rewrite the trinomial as a squared binomial.
6. Simplify.

To solve a quadratic equation by completing the square:

1. Move the constant term to the other side.
2. If the coefficient of \(x^2\) is not 1, divide or factor first.
3. Complete the square.
4. Write the left side as a squared binomial.
5. Take the square root of both sides.
6. Solve for \(x\).

8. Worked Examples

Example 1: Rewrite in vertex form

Rewrite \(y=x^2+4x+1\) in vertex form.

Half of 4 is 2, and \(2^2=4\).

$$y=x^2+4x+4-4+1$$ $$y=(x+2)^2-3$$

Answer:

$$y=(x+2)^2-3$$

The vertex is \((-2,-3)\).

Example 2: Rewrite in vertex form with a leading coefficient not equal to 1

Rewrite \(y=3x^2-12x+7\) in vertex form.

Factor out 3 from the first two terms:

$$y=3(x^2-4x)+7$$

Half of \(-4\) is \(-2\), and \((-2)^2=4\).

$$y=3(x^2-4x+4-4)+7$$ $$y=3((x-2)^2-4)+7$$ $$y=3(x-2)^2-12+7$$ $$y=3(x-2)^2-5$$

Answer:

$$y=3(x-2)^2-5$$

The vertex is \((2,-5)\).

Example 3: Solve by completing the square

Solve \(x^2-10x+21=0\).

Move the constant term:

$$x^2-10x=-21$$

Half of \(-10\) is \(-5\), and \((-5)^2=25\).

$$x^2-10x+25=-21+25$$ $$ (x-5)^2=4 $$

Take square roots:

$$x-5=\pm 2$$

So,

$$x=5\pm 2$$

Therefore,

$$x=7 \quad \text{or} \quad x=3$$

Example 4: Solve when the coefficient of \(x^2\) is not 1

Solve \(2x^2+4x-6=0\).

First divide every term by 2:

$$x^2+2x-3=0$$

Move the constant term:

$$x^2+2x=3$$

Half of 2 is 1, and \(1^2=1\). Add 1 to both sides:

$$x^2+2x+1=3+1$$ $$ (x+1)^2=4 $$

Take square roots:

$$x+1=\pm 2$$

So,

$$x=-1\pm 2$$

Therefore,

$$x=1 \quad \text{or} \quad x=-3$$

9. Common Mistakes to Avoid

• Forgetting to move the constant first when solving an equation.
• Forgetting to add the same number to both sides of an equation.
• Not factoring out the coefficient of \(x^2\) before completing the square when \(a\neq 1\).
• Sign mistakes, especially when halving negative numbers.
• Forgetting the \(\pm\) when taking square roots.

10. Why Completing the Square Matters

This method is important because it connects algebra and graphing. When you rewrite a quadratic in vertex form, you can identify the turning point of the parabola quickly.

It is also one of the main algebra techniques for solving quadratics, especially when factoring is difficult or impossible.

In addition, completing the square helps explain where the quadratic formula comes from.

11. Brief Summary

Completing the square means rewriting a quadratic so part of it becomes a perfect square trinomial.

To do this, take half of the coefficient of \(x\), square it, and then add and subtract that value. This lets you rewrite the quadratic in vertex form or solve a quadratic equation.

With practice, completing the square becomes a reliable method for understanding parabolas and solving quadratic problems.

The Quadratic Formula is a method for solving any quadratic equation written in the form

$$ax^2+bx+c=0$$

where \(a\neq 0\). It is especially useful when a quadratic does not factor easily. In 11th Grade Maths, this formula is important because it helps you find the exact solutions, or roots, of a quadratic equation, including real and complex roots.

A quadratic equation is any equation where the highest power of the variable is 2. For example, \(x^2-5x+6=0\) is quadratic, but \(x^3+2=0\) is not.

The quadratic formula is

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

This formula gives the solutions to any quadratic equation in standard form.

Before using the formula, make sure the equation is in standard form:

• Move all terms to one side so the equation equals 0.
• Identify \(a\), \(b\), and \(c\).
• Substitute carefully into the formula.

For a quadratic equation \(ax^2+bx+c=0\):

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term

For example, in \(3x^2-7x+2=0\):

• \(a=3\)
• \(b=-7\)
• \(c=2\)

Where does the quadratic formula come from? It is derived by completing the square on the general quadratic equation.

Start with

$$ax^2+bx+c=0$$

Since \(a\neq 0\), divide every term by \(a\):

$$x^2+\frac{b}{a}x+\frac{c}{a}=0$$

Move the constant term to the other side:

$$x^2+\frac{b}{a}x=-\frac{c}{a}$$

Now complete the square. Take half of the coefficient of \(x\), which is \(\frac{b}{2a}\), and square it:

$$\left(\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}$$

Add this to both sides:

$$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$$

The left side is now a perfect square trinomial:

$$\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}$$

Write the right side with a common denominator:

$$\left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a^2}$$

Usually this is written as

$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$

Take the square root of both sides:

$$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$

Subtract \(\frac{b}{2a}\) from both sides:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

This is the quadratic formula.

The expression inside the square root, \(b^2-4ac\), is called the discriminant. It tells you what kind of roots the quadratic has.

• If \(b^2-4ac>0\), there are two different real roots.
• If \(b^2-4ac=0\), there is one real repeated root.
• If \(b^2-4ac<0\), there are two complex roots.

This is very useful because you can predict the type of solution before solving the whole equation.

Worked Example 1: A quadratic with two real roots

Solve:

$$x^2-5x+6=0$$

First identify the coefficients:

• \(a=1\)
• \(b=-5\)
• \(c=6\)

Substitute into the quadratic formula:

$$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(6)}}{2(1)}$$

Simplify step by step:

$$x=\frac{5\pm\sqrt{25-24}}{2}$$

$$x=\frac{5\pm\sqrt{1}}{2}$$

$$x=\frac{5\pm 1}{2}$$

Now find the two values:

$$x=\frac{5+1}{2}=3$$

$$x=\frac{5-1}{2}=2$$

So the solutions are

$$x=2 \text{ and } x=3$$

Worked Example 2: A quadratic with one repeated real root

Solve:

$$2x^2+4x+2=0$$

Identify the coefficients:

• \(a=2\)
• \(b=4\)
• \(c=2\)

Use the formula:

$$x=\frac{-4\pm\sqrt{4^2-4(2)(2)}}{2(2)}$$

Simplify:

$$x=\frac{-4\pm\sqrt{16-16}}{4}$$

$$x=\frac{-4\pm\sqrt{0}}{4}$$

$$x=\frac{-4}{4}$$

$$x=-1$$

There is only one solution value, so this quadratic has a repeated root:

$$x=-1$$

The discriminant here is 0, which confirms one repeated real root.

Worked Example 3: A quadratic with irrational real roots

Solve:

$$3x^2+2x-4=0$$

Identify the coefficients:

• \(a=3\)
• \(b=2\)
• \(c=-4\)

Substitute into the formula:

$$x=\frac{-2\pm\sqrt{2^2-4(3)(-4)}}{2(3)}$$

Simplify the discriminant:

$$x=\frac{-2\pm\sqrt{4+48}}{6}$$

$$x=\frac{-2\pm\sqrt{52}}{6}$$

Since \(\sqrt{52}=\sqrt{4\cdot 13}=2\sqrt{13}\), we get

$$x=\frac{-2\pm 2\sqrt{13}}{6}$$

Now simplify by dividing top and bottom by 2:

$$x=\frac{-1\pm \sqrt{13}}{3}$$

So the exact solutions are

$$x=\frac{-1+\sqrt{13}}{3} \text{ and } x=\frac{-1-\sqrt{13}}{3}$$

Worked Example 4: A quadratic with complex roots

Solve:

$$x^2+4x+8=0$$

Identify the coefficients:

• \(a=1\)
• \(b=4\)
• \(c=8\)

Substitute into the formula:

$$x=\frac{-4\pm\sqrt{4^2-4(1)(8)}}{2(1)}$$

Simplify:

$$x=\frac{-4\pm\sqrt{16-32}}{2}$$

$$x=\frac{-4\pm\sqrt{-16}}{2}$$

Since \(\sqrt{-16}=4i\),

$$x=\frac{-4\pm 4i}{2}$$

$$x=-2\pm 2i$$

So the solutions are

$$x=-2+2i \text{ and } x=-2-2i$$

Because the discriminant is negative, the roots are complex.

Tips for using the quadratic formula correctly

• Always write the equation in the form \(ax^2+bx+c=0\) first.
• Be careful with negative signs, especially when substituting for \(b\).
• Use brackets when replacing values in the formula.
• Simplify the square root when possible.
• Give exact answers unless the question asks for decimals.

Common mistakes to avoid

• Forgetting to set the equation equal to 0. For example, if you have \(x^2+3x=10\), first rewrite it as \(x^2+3x-10=0\).
• Using the wrong value of \(b\). In \(x^2-7x+3=0\), \(b=-7\), not 7.
• Making errors with the discriminant. Compute \(b^2-4ac\) carefully.
• Not simplifying fully. For example, \(\sqrt{20}\) should be simplified to \(2\sqrt{5}\).

Why the quadratic formula matters

The quadratic formula works for every quadratic equation. Factoring is often quicker, but not every quadratic factors nicely. Completing the square is useful too, but it can take longer. The quadratic formula is the most reliable general method.

It also connects to the graph of a quadratic, called a parabola. The roots you find are the \(x\)-intercepts of the graph, if the roots are real. If the roots are complex, the parabola does not cross the \(x\)-axis.

Summary

The quadratic formula solves equations of the form \(ax^2+bx+c=0\):

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

It comes from completing the square. The discriminant \(b^2-4ac\) tells you whether the roots are two real roots, one repeated real root, or two complex roots. With careful substitution and simplification, the quadratic formula lets you solve any quadratic equation exactly.

Lesson: The Discriminant and Nature of Roots

When we solve a quadratic equation, we are trying to find the roots or solutions of the equation. A quadratic equation is usually written in the form

$$ax^2+bx+c=0, \quad a\ne 0$$

Sometimes a quadratic has two different real solutions, sometimes one repeated real solution, and sometimes no real solutions at all. The discriminant helps us predict this without fully solving the equation.

The discriminant is the expression inside the square root in the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

So the discriminant is

$$D=b^2-4ac$$

The value of \(D\) tells us the nature of the roots.

Why the discriminant matters

In the quadratic formula, the part that affects the type of answer most is \(\sqrt{b^2-4ac}\). If the number inside the square root is positive, zero, or negative, the solutions behave differently.

• If \(D>0\), the quadratic has two distinct real roots.
• If \(D=0\), the quadratic has one repeated real root.
• If \(D<0\), the quadratic has two complex conjugate roots, which means there are no real roots.

This is the most important classification to remember.

Going further: rational or irrational roots

If \(D>0\), we know there are two real roots. But we can often say more.

• If \(D\) is a perfect square (such as 1, 4, 9, 16, 25), then the roots are usually rational.
• If \(D\) is not a perfect square, then the roots are usually irrational.

For example, if \(D=36\), then \(\sqrt{D}=6\), which is rational. But if \(D=20\), then \(\sqrt{20}=2\sqrt{5}\), which is irrational.

So we can summarize the nature of roots like this:

• \(D>0\) and perfect square: two distinct real rational roots
• \(D>0\) and not a perfect square: two distinct real irrational roots
• \(D=0\): one repeated real rational root
• \(D<0\): two non-real complex conjugate roots

Connection to the graph

The graph of a quadratic equation is a parabola. The discriminant also tells us how the parabola meets the \(x\)-axis.

• If \(D>0\), the parabola crosses the \(x\)-axis at two points.
• If \(D=0\), the parabola touches the \(x\)-axis at one point.
• If \(D<0\), the parabola does not meet the \(x\)-axis.

This makes sense because the roots are the \(x\)-values where the graph equals zero.

Step-by-step method

1. Write the quadratic in the form \(ax^2+bx+c=0\).
2. Identify \(a\), \(b\), and \(c\).
3. Substitute into \(D=b^2-4ac\).
4. Calculate the value of \(D\).
5. Use the sign and type of \(D\) to classify the roots.

Worked Example 1: Two distinct real rational roots

Classify the roots of

$$x^2-5x+6=0$$

Here, \(a=1\), \(b=-5\), and \(c=6\).

Find the discriminant:

$$D=b^2-4ac=(-5)^2-4(1)(6)=25-24=1$$

Since \(D=1\), we know:

• \(D>0\), so there are two distinct real roots.
• \(1\) is a perfect square, so the roots are rational.

So this quadratic has two distinct real rational roots.

In fact, if we solve it, the roots are \(x=2\) and \(x=3\).

Worked Example 2: Two distinct real irrational roots

Classify the roots of

$$2x^2+3x-4=0$$

Here, \(a=2\), \(b=3\), and \(c=-4\).

Find the discriminant:

$$D=b^2-4ac=3^2-4(2)(-4)=9+32=41$$

Since \(D=41\), we know:

• \(D>0\), so there are two distinct real roots.
• \(41\) is not a perfect square, so the roots are irrational.

So this quadratic has two distinct real irrational roots.

Worked Example 3: One repeated real root

Classify the roots of

$$4x^2-12x+9=0$$

Here, \(a=4\), \(b=-12\), and \(c=9\).

Find the discriminant:

$$D=b^2-4ac=(-12)^2-4(4)(9)=144-144=0$$

Since \(D=0\), the quadratic has one repeated real root.

This means both roots are the same value. If solved, the answer is

$$x=\frac{-(-12)}{2(4)}=\frac{12}{8}=\frac{3}{2}$$

So the equation has a repeated real root at \(x=\frac{3}{2}\).

Worked Example 4: Complex conjugate roots

Classify the roots of

$$x^2+4x+8=0$$

Here, \(a=1\), \(b=4\), and \(c=8\).

Find the discriminant:

$$D=b^2-4ac=4^2-4(1)(8)=16-32=-16$$

Since \(D<0\), there are no real roots.

Instead, the equation has two complex conjugate roots.

If solved using the quadratic formula:

$$x=\frac{-4\pm\sqrt{-16}}{2}=\frac{-4\pm 4i}{2}=-2\pm 2i$$

So the roots are \(-2+2i\) and \(-2-2i\).

Important notes and common mistakes

• Always write the equation in standard form \(ax^2+bx+c=0\) before identifying \(a\), \(b\), and \(c\).
• Be careful with negative signs, especially when substituting \(b\) and \(c\).
• Remember that \(b^2\) means the whole value of \(b\) is squared. For example, if \(b=-7\), then \(b^2=(-7)^2=49\).
• Do not confuse \(D=0\) with no roots. It means one repeated real root.
• If \(D<0\), the equation still has solutions, but they are complex, not real.

Using the discriminant without solving fully

One of the most useful things about the discriminant is that it lets you classify the roots quickly. This is especially helpful when a question asks only for the nature of the roots, not the actual roots.

For example, for

$$3x^2-6x+5=0$$

we compute

$$D=(-6)^2-4(3)(5)=36-60=-24$$

Since \(D<0\), we can immediately say: the equation has two complex conjugate roots and no real roots.

Quick classification chart

• \(D>0\): two different real roots
• \(D=0\): one repeated real root
• \(D<0\): two complex conjugate roots

And if \(D>0\):

• perfect square \(\Rightarrow\) rational roots
• not a perfect square \(\Rightarrow\) irrational roots

Brief Summary

The discriminant of a quadratic equation \(ax^2+bx+c=0\) is \(D=b^2-4ac\). It tells us whether the roots are real, repeated, or complex. A positive discriminant gives two distinct real roots, zero gives one repeated real root, and a negative discriminant gives two complex conjugate roots. If the discriminant is positive, checking whether it is a perfect square tells us whether the real roots are rational or irrational.

Parabolic Geometry and Vertex Form

Quadratic relations create graphs called parabolas. These graphs appear in many situations, such as the path of a thrown ball, the shape of a satellite dish, or the curve of a bridge arch. To understand a parabola well, we need to recognize its key geometric features and know how its equation affects its shape and position.

One of the most useful ways to write a quadratic is vertex form. Vertex form makes it easier to identify the turning point of the parabola and describe how the graph has been transformed from the parent quadratic.

The parent quadratic is

$$y=x^2$$

This parabola opens upward, has a vertex at \,(0,0)\, and is symmetric about the y-axis.

The vertex form of a quadratic is

$$y=a(x-h)^2+k$$

Each part of this equation tells us something important about the graph.

• Vertex: \,(h,k)\,
• Axis of symmetry: the vertical line \,x=h\,
• 1a value: controls whether the parabola opens up or down, and how wide or narrow it is

Here is what the value of \,a\, tells us:

• If \,a>0\,, the parabola opens upward.
• If \,a<0\,, the parabola opens downward.
• If \,|a|>1\,, the parabola is narrower than \,y=x^2\,.
• If \,0<|a|<1\,, the parabola is wider than \,y=x^2\,.

The numbers \,h\, and \,k\, describe translations, or shifts, of the graph.

• \,h\, shifts the graph left or right.
• \,k\, shifts the graph up or down.

Be careful with the sign of \,h\,. In the expression \,(x-h)^2\,, the graph shifts right if \,h\, is positive and left if \,h\, is negative.

For example:

• \,y=(x-3)^2+2\, has vertex \,(3,2)\,
• \,y=(x+4)^2-1\, has vertex \,(-4,-1)\,

Key geometric features of a parabola

When graphing or analyzing a parabola, these are the main features to identify:

• Vertex: the highest or lowest point on the parabola
• Axis of symmetry: the vertical line passing through the vertex
• x-intercepts: points where the graph crosses the x-axis, so \,y=0\,
• y-intercept: the point where the graph crosses the y-axis, so \,x=0\,
• Opening: whether the parabola opens up or down

The vertex is the minimum point if the parabola opens upward, and the maximum point if it opens downward.

How to graph a parabola from vertex form

1. Find the vertex \,(h,k)\,.
2. Draw the axis of symmetry \,x=h\,.
3. Determine whether the parabola opens up or down from the sign of \,a\,.
4. Use the value of \,a\, to judge whether it is wider or narrower than \,y=x^2\,.
5. Find additional points by substituting x-values near the vertex.
6. Reflect points across the axis of symmetry.
7. Find intercepts if needed.

Finding intercepts

To find the y-intercept, substitute \,x=0\, into the equation.

To find the x-intercepts, set \,y=0\, and solve the quadratic equation. A parabola may have:

• two x-intercepts
• one x-intercept if the vertex lies on the x-axis
• no x-intercepts if the graph does not cross the x-axis

Transformations from the parent quadratic

Vertex form is especially useful because it shows transformations clearly. Starting from \,y=x^2\,:

• 1a stretch or compression changes the width through \,a\,
• reflection across the x-axis happens when \,a\, is negative
• horizontal shift comes from \,h\,
• vertical shift comes from \,k\,

For example, in

$$y=-2(x-1)^2+3$$

• the graph shifts right 1
• the graph shifts up 3
• it reflects across the x-axis because of the negative sign
• it becomes narrower because \,|a|=2>1\,

Worked Example 1: Identify features from vertex form

Find the vertex, axis of symmetry, opening, and y-intercept of

$$y=(x-2)^2-5$$

Step 1: Find the vertex.

The equation is already in vertex form \,y=a(x-h)^2+k\,. Here, \,h=2\, and \,k=-5\,. So the vertex is

$$ (2,-5) $$

Step 2: Find the axis of symmetry.

The axis of symmetry is

$$x=2$$

Step 3: Determine the opening.

Since \,a=1>0\,, the parabola opens upward.

Step 4: Find the y-intercept.

Substitute \,x=0\,:

$$y=(0-2)^2-5=4-5=-1$$

So the y-intercept is

$$ (0,-1) $$

Result:

• Vertex: \,(2,-5)\,
• Axis of symmetry: \,x=2\,
• Opens upward
• y-intercept: \,(0,-1)\,

Worked Example 2: Find the x-intercepts

Find the x-intercepts of

$$y=(x+1)^2-4$$

For x-intercepts, set \,y=0\,:

$$0=(x+1)^2-4$$

Add 4 to both sides:

$$4=(x+1)^2$$

Take the square root of both sides:

$$x+1=\pm 2$$

Now solve both cases:

$$x+1=2 \Rightarrow x=1$$

$$x+1=-2 \Rightarrow x=-3$$

So the x-intercepts are

$$ (1,0) \text{ and } (-3,0) $$

Notice that the vertex is \,(-1,-4)\,. The axis of symmetry is \,x=-1\,, which lies exactly halfway between \,x=1\, and \,x=-3\,. This is a useful check.

Worked Example 3: Describe transformations and graph features

Analyze the parabola

$$y=-\frac{1}{2}(x-4)^2+1$$

Step 1: Identify the vertex.

The vertex is

$$ (4,1) $$

Step 2: Axis of symmetry.

$$x=4$$

Step 3: Opening and width.

Since \,a=-\frac{1}{2}\,, the parabola opens downward. Because \,|a|<1\,, it is wider than \,y=x^2\,.

Step 4: Describe the transformations.

• shift right 4
• shift up 1
• reflect across the x-axis
• vertical compression by a factor of \,\frac{1}{2}\,

Step 5: Find the y-intercept.

Substitute \,x=0\,:

$$y=-\frac{1}{2}(0-4)^2+1=-\frac{1}{2}(16)+1=-8+1=-7$$

So the y-intercept is

$$ (0,-7) $$

Step 6: Find the x-intercepts.

Set \,y=0\,:

$$0=-\frac{1}{2}(x-4)^2+1$$

Subtract 1:

$$-1=-\frac{1}{2}(x-4)^2$$

Multiply both sides by \, -2\,:

$$2=(x-4)^2$$

Take the square root:

$$x-4=\pm \sqrt{2}$$

So

$$x=4\pm \sqrt{2}$$

The x-intercepts are

$$ (4-\sqrt{2},0) \text{ and } (4+\sqrt{2},0) $$

Worked Example 4: Write an equation from graph information

A parabola has vertex \,(3,-2)\, and opens upward with the same width as \,y=x^2\,. Write its equation in vertex form.

Use the pattern

$$y=a(x-h)^2+k$$

Since the vertex is \,(3,-2)\,, we have \,h=3\, and \,k=-2\,.

Because it opens upward with the same width as the parent quadratic, \,a=1\,.

So the equation is

$$y=(x-3)^2-2$$

If the parabola had opened downward with the same width, then \,a=-1\, and the equation would be

$$y=-(x-3)^2-2$$

Helpful graphing strategy

Once you know the vertex, choose x-values one unit to the left and right of the vertex. Then choose x-values two units to the left and right. Because a parabola is symmetric, points on one side match points on the other side.

For example, for \,y=(x-2)^2-1\,:

• vertex: \,(2,-1)\,
• at \,x=3\,, \,y=(3-2)^2-1=0\,, so \,(3,0)\,
• by symmetry, \,(1,0)\, is also on the graph
• at \,x=4\,, \,y=(4-2)^2-1=3\,, so \,(4,3)\,
• by symmetry, \,(0,3)\, is also on the graph

This gives enough points to sketch a smooth parabola.

Common mistakes to avoid

• Sign error with the vertex: in \,y=(x-5)^2+2\,, the vertex is \,(5,2)\,, not \,(-5,2)\,.
• Forgetting the axis of symmetry: it is always \,x=h\, in vertex form.
• Mixing up opening and shifting: \,a\, changes opening and width, while \,h\, and \,k\, shift the graph.
• Assuming every parabola has x-intercepts: some do not cross the x-axis.

Summary

Vertex form, \,y=a(x-h)^2+k\,, is a powerful way to understand a parabola quickly. It tells you the vertex \,(h,k)\,, the axis of symmetry \,x=h\,, whether the parabola opens up or down, and whether it is wider or narrower than the parent graph.

By using vertex form, you can graph parabolas, find intercepts, and describe transformations from \,y=x^2\,. When solving problems, always identify the vertex first, then use symmetry and intercepts to build the rest of the graph.

Vieta's Formulas for Quadratics

When we solve a quadratic equation, we often look for its roots, also called its solutions or zeros. For example, in the equation \(x^2-5x+6=0\), the roots are \(2\) and \(3\).

Vieta's Formulas give a very useful connection between the roots of a quadratic and the coefficients of the equation. This means we can learn about the roots without fully solving the equation every time.

This idea is especially helpful when:

• checking whether your answers make sense,
• building a quadratic equation from known roots,
• finding unknown numbers when only the sum or product of roots is given,
• solving worded algebra problems more efficiently.

1. The standard form of a quadratic

A quadratic equation is usually written in the form

$$ax^2+bx+c=0$$

where \(a\), \(b\), and \(c\) are numbers and \(a\neq 0\).

Suppose the roots of this equation are \(r_1\) and \(r_2\). Vieta's Formulas tell us:

$$r_1+r_2=-\frac{b}{a}$$ $$r_1r_2=\frac{c}{a}$$

So:

• the sum of the roots is \(-\frac{b}{a}\),
• the product of the roots is \(\frac{c}{a}\).

2. Why these formulas are true

If a quadratic has roots \(r_1\) and \(r_2\), then it can be written in factored form as

$$a(x-r_1)(x-r_2)=0$$

Now expand the factors:

$$a(x-r_1)(x-r_2)=a\left(x^2-(r_1+r_2)x+r_1r_2\right)$$

Distribute \(a\):

$$ax^2-a(r_1+r_2)x+a(r_1r_2)$$

Compare this with the standard form

$$ax^2+bx+c$$

Matching the coefficients gives:

• \(-a(r_1+r_2)=b\), so \(r_1+r_2=-\frac{b}{a}\)
• \(a(r_1r_2)=c\), so \(r_1r_2=\frac{c}{a}\)

This is the main idea behind Vieta's Formulas.

3. Important note about signs

Students often make sign mistakes with Vieta's Formulas. The two most important facts to remember are:

• the sum of the roots uses the opposite sign of \(b\),
• the product of the roots keeps the same sign pattern as \(c\) divided by \(a\).

For \(ax^2+bx+c=0\):

• Sum \(= -\frac{b}{a}\)
• Product \(= \frac{c}{a}\)

4. Worked Example 1: Find the sum and product of the roots

Find the sum and product of the roots of

$$x^2-7x+10=0$$

Here, \(a=1\), \(b=-7\), and \(c=10\).

Use Vieta's Formulas:

$$r_1+r_2=-\frac{b}{a}=-\frac{-7}{1}=7$$ $$r_1r_2=\frac{c}{a}=\frac{10}{1}=10$$

So the roots have:

• sum \(7\)
• product \(10\)

We can check by factoring:

$$x^2-7x+10=(x-5)(x-2)$$

The roots are \(5\) and \(2\). Indeed:

$$5+2=7 \quad \text{and} \quad 5\cdot 2=10$$

5. Worked Example 2: Use the sum and product to identify the roots

Find the roots of

$$x^2+x-12=0$$

First identify the coefficients:

• \(a=1\)
• \(b=1\)
• \(c=-12\)

Now apply Vieta's Formulas:

$$r_1+r_2=-\frac{1}{1}=-1$$ $$r_1r_2=\frac{-12}{1}=-12$$

We now ask: which two numbers add to \(-1\) and multiply to \(-12\)?

Those numbers are \(3\) and \(-4\), because

$$3+(-4)=-1$$ $$3\cdot(-4)=-12$$

So the roots are

$$x=3 \quad \text{and} \quad x=-4$$

This matches the factorization:

$$x^2+x-12=(x+4)(x-3)$$

6. Worked Example 3: Quadratic with leading coefficient not equal to 1

Find the sum and product of the roots of

$$2x^2-9x+4=0$$

Here:

• \(a=2\)
• \(b=-9\)
• \(c=4\)

Use the formulas:

$$r_1+r_2=-\frac{-9}{2}=\frac{9}{2}$$ $$r_1r_2=\frac{4}{2}=2$$

So the sum of the roots is \(\frac{9}{2}\), and the product is \(2\).

If we solve the equation, we can verify this result. Factor:

$$2x^2-9x+4=(2x-1)(x-4)$$

The roots are \(x=\frac{1}{2}\) and \(x=4\).

Check:

$$\frac{1}{2}+4=\frac{1}{2}+\frac{8}{2}=\frac{9}{2}$$ $$\frac{1}{2}\cdot 4=2$$

7. Worked Example 4: Form a quadratic from given roots

Form a quadratic equation whose roots are \(3\) and \(-5\).

If the roots are \(3\) and \(-5\), then the factors are:

$$ (x-3)(x+5) $$

Expand:

$$ (x-3)(x+5)=x^2+2x-15 $$

So the quadratic equation is

$$x^2+2x-15=0$$

Now check using Vieta's Formulas:

• sum of roots: \(3+(-5)=-2\)
• product of roots: \(3\cdot(-5)=-15\)

In the equation \(x^2+2x-15=0\), we have:

$$-\frac{b}{a}=-\frac{2}{1}=-2$$ $$\frac{c}{a}=\frac{-15}{1}=-15$$

The values match perfectly.

8. Using Vieta's Formulas in reverse

Sometimes you are told the sum and product of two numbers and asked to form a quadratic whose roots are those numbers.

If the roots are \(r_1\) and \(r_2\), then:

• sum \(=S=r_1+r_2\)
• product \(=P=r_1r_2\)

The quadratic with those roots is

$$x^2-Sx+P=0$$

This works when the leading coefficient is \(1\).

For example, if the sum of the roots is \(8\) and the product is \(12\), then:

$$x^2-8x+12=0$$

Its roots are \(2\) and \(6\), since

$$2+6=8 \quad \text{and} \quad 2\cdot 6=12$$

9. Connection to solving quadratics

Vieta's Formulas do not replace methods like factoring, completing the square, or the quadratic formula. Instead, they help you understand the structure of the equation.

For example, if you solve a quadratic and get two answers, you can quickly check them:

• Do they add to \(-\frac{b}{a}\)?
• Do they multiply to \(\frac{c}{a}\)?

If not, there is likely an error.

10. Common mistakes to avoid

• Forgetting the negative sign in the sum formula. The sum is \(-\frac{b}{a}\), not \(\frac{b}{a}\).
• Ignoring the value of \(a\). If \(a\neq 1\), you must divide by \(a\).
• Mixing up sum and product. Sum comes from \(b\); product comes from \(c\).
• Confusing roots with factors. If \((x-2)=0\), then the root is \(2\), not \(-2\).

11. Quick practice ideas

Try these on your own:

1. For \(x^2-4x-21=0\), find the sum and product of the roots.
2. Find two numbers with sum \(6\) and product \(8\), then form the quadratic.
3. For \(3x^2+2x-1=0\), find the sum and product of the roots.

Answers:

• 1. Sum \(=4\), product \(=-21\)
• 2. The numbers are \(2\) and \(4\), so the quadratic is \(x^2-6x+8=0\)
• 3. Sum \(=-\frac{2}{3}\), product \(=-\frac{1}{3}\)

12. Summary

Vieta's Formulas connect the roots of a quadratic equation directly to its coefficients. For \(ax^2+bx+c=0\) with roots \(r_1\) and \(r_2\):

$$r_1+r_2=-\frac{b}{a}, \qquad r_1r_2=\frac{c}{a}$$

These formulas help you find the sum and product of roots, check solutions, and create quadratic equations from known roots. Once you are comfortable with the signs and the role of \(a\), Vieta's Formulas become a fast and powerful algebra tool.

Quadratic Inequalities are inequalities that involve a quadratic expression, such as:

\(x^2-5x+6>0\), \(2x^2+3x-2\le 0\), or even inequalities in two variables like \(y<x^2-4x+3\).

In this lesson, you will learn how to solve quadratic inequalities in one variable and how to understand quadratic inequalities in two variables using graphs. The main idea is that a quadratic expression can change sign depending on the value of the variable, so we need to find where it is positive, negative, or zero.

1. What makes a quadratic inequality different from a quadratic equation?

A quadratic equation asks for exact values where the expression equals zero, such as:

$$x^2-5x+6=0$$

A quadratic inequality asks where the expression is greater than, less than, greater than or equal to, or less than or equal to zero, such as:

$$x^2-5x+6>0$$

Instead of getting just one or two answers, the solution is often an interval or a union of intervals.

2. The basic strategy for solving quadratic inequalities in one variable

To solve a quadratic inequality, use these steps:

1. Move everything to one side so the inequality is compared to zero.
2. Factor the quadratic if possible.
3. Find the critical values, which are the values that make the expression equal zero.
4. Use a sign chart or test values to determine where the expression is positive or negative.
5. Write the solution in interval form or inequality form.

3. Why sign charts work

A factored quadratic, such as \((x-2)(x-3)\), changes sign only at its zeros. Those zeros split the number line into intervals. Inside each interval, the sign of the expression stays the same.

So once we know the zeros, we can test one value from each interval to find whether the quadratic is positive or negative there.

4. Worked Example 1: Solving a simple factored inequality

Solve:

$$x^2-5x+6>0$$

Step 1: Factor

$$x^2-5x+6=(x-2)(x-3)$$

So we solve:

$$ (x-2)(x-3)>0 $$

Step 2: Find the critical values

The expression is zero when:

$$x=2 \quad \text{or} \quad x=3$$

These values divide the number line into three intervals:

• \(( -\infty,2 )\)
• \((2,3)\)
• \((3,\infty)\)

Step 3: Test each interval

Pick a value from each interval:

• If \(x=1\), then \((1-2)(1-3)=(-1)(-2)=2\), which is positive.
• If \(x=2.5\), then \((2.5-2)(2.5-3)=(0.5)(-0.5)=-0.25\), which is negative.
• If \(x=4\), then \((4-2)(4-3)=(2)(1)=2\), which is positive.

Step 4: Choose the intervals that satisfy the inequality

We want the expression to be greater than zero, so we take the positive intervals:

$$x<2 \quad \text{or} \quad x>3$$

In interval form:

$$(-\infty,2)\cup(3,\infty)$$

Notice that \(2\) and \(3\) are not included because the inequality is strict: \(>0\), not \(\ge 0\).

5. Worked Example 2: Including endpoints

Solve:

$$2x^2+3x-2\le 0$$

Step 1: Factor

$$2x^2+3x-2=(2x-1)(x+2)$$

So we solve:

$$ (2x-1)(x+2)\le 0 $$

Step 2: Find the critical values

Set each factor equal to zero:

$$2x-1=0 \Rightarrow x=\frac12$$

$$x+2=0 \Rightarrow x=-2$$

The intervals are:

• \(( -\infty,-2 )\)
• \(( -2,\frac12 )\)
• \(( \frac12,\infty )\)

Step 3: Test each interval

• If \(x=-3\), then \((2x-1)(x+2)=(-7)(-1)=7\), positive.
• If \(x=0\), then \((2(0)-1)(0+2)=(-1)(2)=-2\), negative.
• If \(x=1\), then \((2(1)-1)(1+2)=(1)(3)=3\), positive.

Step 4: Choose the correct interval

We need the expression to be less than or equal to zero. That means:

• Take the interval where the expression is negative.
• Include the points where the expression equals zero.

So the solution is:

$$-2\le x\le \frac12$$

In interval form:

$$[-2,\tfrac12]$$

6. A quick pattern to remember

For a quadratic with two distinct real roots and a positive leading coefficient, the parabola opens upward.

• It is positive outside the roots.
• It is negative between the roots.

For a quadratic with a negative leading coefficient, the parabola opens downward.

• It is negative outside the roots.
• It is positive between the roots.

This pattern is helpful, but you should still be ready to confirm using factoring, a sign chart, or the graph.

7. Solving when the quadratic is not already factored

Sometimes the expression must be rearranged first, and then factored.

Worked Example 3: Rearranging first

Solve:

$$x^2+4x<5$$

Step 1: Move everything to one side

$$x^2+4x-5<0$$

Step 2: Factor

$$x^2+4x-5=(x+5)(x-1)$$

So we solve:

$$ (x+5)(x-1)<0 $$

Step 3: Find the critical values

$$x=-5 \quad \text{and} \quad x=1$$

Step 4: Test intervals

• If \(x=-6\), then \((x+5)(x-1)=(-1)(-7)=7\), positive.
• If \(x=0\), then \((5)(-1)=-5\), negative.
• If \(x=2\), then \((7)(1)=7\), positive.

Step 5: Select the interval

We want the expression to be less than zero, so:

$$-5<x<1$$

In interval form:

$$(-5,1)$$

8. What if the quadratic cannot be factored easily?

In Grade 11, many questions are chosen so that factoring works nicely. But if a quadratic does not factor easily, you can still:

• find the roots using another method you know,
• then use those roots in a sign chart,
• or use the graph of the parabola.

The important idea is always the same: find where the expression is zero, then decide where it is above or below zero.

9. Solving using the graph of a parabola

A quadratic expression like \(y=x^2-5x+6\) graphs as a parabola.

To solve an inequality such as:

$$x^2-5x+6>0$$

you are asking: For which x-values is the graph above the x-axis?

Since the graph crosses the x-axis at \(x=2\) and \(x=3\), the parabola is above the x-axis when:

$$x<2 \quad \text{or} \quad x>3$$

This matches the answer from the sign chart method.

10. Special case: a repeated root

Sometimes a quadratic touches the x-axis at only one point. For example:

$$x^2-4x+4\ge 0$$

Factor:

$$x^2-4x+4=(x-2)^2$$

A square is never negative, so \((x-2)^2\ge 0\) for all real x.

Therefore, the solution is:

$$(-\infty,\infty)$$

But if the inequality were:

$$x^2-4x+4<0$$

then there would be no solution, because a square cannot be less than zero.

11. Quadratic inequalities in two variables

Now let us look at inequalities such as:

$$y<x^2-4x+3$$

This is different from a one-variable inequality because the solution is not a set of points on a number line. Instead, the solution is a region on a graph.

First graph the boundary:

$$y=x^2-4x+3$$

This is a parabola opening upward.

Then decide whether the solution is above or below the parabola:

• \(y>x^2-4x+3\) means points above the parabola.
• \(y<x^2-4x+3\) means points below the parabola.

Because the inequality is strict, \(<\) or \(>\), the boundary parabola is usually drawn dashed, meaning points on the curve are not included.

If the inequality is \(\le\) or \(\ge\), the boundary is drawn solid, because points on the curve are included.

Worked Example 4: Interpreting a quadratic inequality in two variables

Describe the solution to:

$$y\ge x^2-2x-3$$

Step 1: Identify the boundary

The boundary is the parabola:

$$y=x^2-2x-3$$

Step 2: Decide on solid or dashed boundary

Because the symbol is \(\ge\), the boundary is solid.

Step 3: Decide which region to shade

Since \(y\) is greater than or equal to the quadratic expression, the solution is the region above the parabola, including the parabola itself.

Optional check using a test point

Test the point \((0,0)\):

$$0\ge 0^2-2(0)-3$$

$$0\ge -3$$

This is true, so \((0,0)\) is part of the solution region. If your graph is correct, the shaded region should include \((0,0)\).

12. Common mistakes to avoid

• Forgetting to move everything to one side. It is much easier to compare the quadratic to zero.
• Using roots but not checking signs. The roots alone are not the full answer; you must decide which intervals work.
• Including endpoints when the inequality is strict. For \(>\) or \(<\), do not include the roots.
• Leaving out endpoints for \(\ge\) or \(\le\). If the expression can equal zero, include those values.
• Shading the wrong side of a parabola in two variables. Use a test point if you are unsure.

13. Quick checklist for one-variable quadratic inequalities

• Rewrite the inequality so one side is zero.
• Factor the quadratic if possible.
• Find the zeros.
• Split the number line into intervals.
• Use test values or a sign chart.
• Write the answer carefully, including or excluding endpoints correctly.

14. Quick checklist for two-variable quadratic inequalities

• Graph the related equation as the boundary.
• Use a dashed boundary for \(<\) or \(>\).
• Use a solid boundary for \(\le\) or \(\ge\).
• Shade above or below the parabola based on the inequality symbol.
• Check with a test point if needed.

Summary

To solve a quadratic inequality in one variable, first rewrite it so one side is zero, then factor, find the zeros, and use a sign chart or test points to determine where the expression is positive or negative.

For inequalities in two variables, graph the parabola as a boundary and shade the region above or below it depending on the inequality symbol. Always pay attention to whether the boundary should be solid or dashed.

Quadratic Optimization is the process of using a quadratic function to find a maximum or minimum value in a real situation.

In Grade 11, this often comes up when a problem involves area, profit, height, distance, or another quantity that changes in a curved way. Since quadratic graphs are parabolas, they have a turning point called the vertex. That vertex tells us the extreme value.

If a parabola opens downward, the vertex gives the maximum value. If it opens upward, the vertex gives the minimum value.

This lesson will show you how to:

• recognize when a situation can be modeled by a quadratic function,
• write the quadratic equation,
• find the vertex,
• interpret the vertex in context, and
• check that your answer makes sense.

1. Review: What is a quadratic?

A quadratic function can be written in standard form as

$$y=ax^2+bx+c$$

where:

• \(a\neq 0\)

• the graph is a parabola,

• the parabola has a vertex, which is its highest or lowest point.

The direction of opening depends on the value of \(a\):

• If \(a>0\), the parabola opens upward, so the vertex is a minimum.

• If \(a<0\), the parabola opens downward, so the vertex is a maximum.

2. Why the vertex matters in optimization

Optimization means finding the best possible value. In many quadratic problems, “best” means largest or smallest.

For a quadratic function, the best value happens at the vertex. So if you can find the vertex, you can solve the optimization problem.

The x-coordinate of the vertex is given by

$$x=-\frac{b}{2a}$$

Then substitute that value back into the quadratic function to find the y-coordinate.

So for

$$y=ax^2+bx+c$$

the steps are:

1. Identify \(a\), \(b\), and \(c\).

2. Find the x-coordinate of the vertex using \(x=-\frac{b}{2a}\).

3. Substitute into the equation to find the y-coordinate.

4. State what the result means in the context of the problem.

3. Forms of a quadratic and how they help

Quadratics can appear in different forms.

Standard form:

$$y=ax^2+bx+c$$

This form is useful because you can quickly use

$$x=-\frac{b}{2a}$$

to find the vertex.

Vertex form:

$$y=a(x-h)^2+k$$

In this form, the vertex is directly visible as \((h,k)\).

This is especially useful in optimization because:

• \(h\) tells where the maximum or minimum happens,

• \(k\) tells the maximum or minimum value.

4. Interpreting the vertex in context

When solving optimization problems, the vertex is not just a point on a graph. It has a real meaning.

For example:

• In an area problem, the vertex might represent the dimensions that create the greatest area.

• In a projectile problem, the vertex might represent the highest point reached.

• In a cost problem, the vertex might represent the smallest possible cost.

Always include units and explain what the numbers mean.

5. Building a quadratic model from a situation

Some questions already give you a quadratic equation. Others require you to create one from a word problem.

Here are two common situations:

A. Rectangle area maximization

Suppose a rectangle has a fixed perimeter. If one side is \(x\), the other side can be written in terms of \(x\). Then area becomes a quadratic function.

Example idea:

• If perimeter is 40 m, then \(2l+2w=40\), so \(l+w=20\).

• If width is \(x\), then length is \(20-x\).

• Area is \(A=x(20-x)=20x-x^2\).

That area function is quadratic, so the maximum area occurs at the vertex.

B. Projectile motion

The height of a projectile is often modeled by a quadratic equation in time:

$$h(t)=at^2+bt+c$$

Because the object rises and then falls, the parabola usually opens downward. That means the vertex gives the maximum height.

6. Worked Example 1: Finding a maximum from a given quadratic

A ball’s height is modeled by

$$h(t)=-5t^2+20t+3$$

where \(h\) is in metres and \(t\) is in seconds.

Find the maximum height.

Step 1: Identify \(a\) and \(b\)

Here, \(a=-5\) and \(b=20\).

Step 2: Find the time at the vertex

$$t=-\frac{b}{2a}=-\frac{20}{2(-5)}=-\frac{20}{-10}=2$$

So the ball reaches its maximum height at \(t=2\) seconds.

Step 3: Find the height at that time

$$h(2)=-5(2)^2+20(2)+3$$ $$h(2)=-5(4)+40+3$$ $$h(2)=-20+40+3=23$$

Answer: The maximum height is 23 m, reached after 2 s.

Why this makes sense: The coefficient of \(t^2\) is negative, so the parabola opens downward. That means the vertex must be a maximum.

7. Worked Example 2: Maximizing area

A farmer has 60 m of fencing to build a rectangular pen. What dimensions will give the maximum area?

Step 1: Write a perimeter equation

$$2l+2w=60$$

Divide by 2:

$$l+w=30$$

Let width be \(x\). Then length is

$$30-x$$

Step 2: Write the area function

$$A=x(30-x)$$ $$A=30x-x^2$$

This is a quadratic:

$$A=-x^2+30x$$

Step 3: Find the vertex

Here, \(a=-1\) and \(b=30\).

$$x=-\frac{b}{2a}=-\frac{30}{2(-1)}=15$$

So one dimension is 15 m.

The other dimension is

$$30-15=15$$

Step 4: Find the maximum area

$$A(15)=30(15)-(15)^2$$ $$A(15)=450-225=225$$

Answer: The maximum area is 225 m², and it happens when the pen is 15 m by 15 m.

Important idea: For a fixed perimeter, the rectangle with the greatest area is a square.

8. Worked Example 3: Minimizing a quadratic

The cost of producing \(x\) items is modeled by

$$C(x)=2x^2-16x+50$$

Find the minimum cost.

Step 1: Decide whether there is a maximum or minimum

Since \(a=2>0\), the parabola opens upward. So the vertex gives a minimum.

Step 2: Find the x-coordinate of the vertex

$$x=-\frac{b}{2a}=-\frac{-16}{2(2)}=\frac{16}{4}=4$$

Step 3: Find the minimum cost

$$C(4)=2(4)^2-16(4)+50$$ $$C(4)=2(16)-64+50$$ $$C(4)=32-64+50=18$$

Answer: The minimum cost is 18 (units of currency), and it occurs when 4 items are produced.

9. Worked Example 4: Using vertex form in context

The height of a firework is given by

$$h(t)=-4(t-3)^2+40$$

Find:

• the maximum height,

• when it occurs.

Step 1: Recognize the form

This is vertex form:

$$h(t)=a(t-h)^2+k$$

So the vertex is \((3,40)\).

Step 2: Interpret the vertex

The firework reaches its highest point at 3 s, and the maximum height is 40 m.

Answer: Maximum height = 40 m; time of maximum height = 3 s.

10. Completing the square and optimization

Sometimes you may want to rewrite a quadratic from standard form into vertex form. This helps you see the maximum or minimum more clearly.

Example:

$$y=x^2-6x+11$$

Complete the square:

$$y=(x^2-6x+9)+2$$ $$y=(x-3)^2+2$$

Now the vertex is \((3,2)\).

Since the coefficient of \((x-3)^2\) is positive, the parabola opens upward. So the minimum value is 2, and it occurs when \(x=3\).

11. Common mistakes to avoid

• Forgetting what the variables mean. In word problems, \(x\) may represent width, time, or number of items. Always define it.

• Finding only the x-coordinate of the vertex. You usually need both the x-value and the y-value.

• Not checking if the answer is a maximum or minimum. Look at the sign of \(a\).

• Ignoring the context. A negative length or negative time may not make sense in a real situation.

• Leaving the answer without units. State metres, seconds, square metres, dollars, and so on.

12. Problem-solving strategy for quadratic optimization

When you see a word problem, use this plan:

1. Decide what quantity you want to maximize or minimize.

2. Choose a variable and define it clearly.

3. Write an equation for the quantity in terms of that variable.

4. Simplify until you have a quadratic function.

5. Find the vertex using \(x=-\frac{b}{2a}\) or vertex form.

6. Interpret the result in the situation.

7. Check whether the answer is reasonable.

13. Quick practice ideas

Try asking yourself these questions when solving:

• What does the vertex represent here?

• Does the parabola open up or down?

• Am I finding the maximum value, minimum value, or the input where it happens?

• Did I substitute back into the function?

14. Summary

Quadratic optimization uses the fact that every quadratic has a vertex, which is the highest or lowest point of the parabola.

For a quadratic in standard form,

$$y=ax^2+bx+c$$

the x-coordinate of the vertex is

$$x=-\frac{b}{2a}$$

If \(a<0\), the vertex gives a maximum. If \(a>0\), the vertex gives a minimum.

In real-world problems, you first build a quadratic model, then use the vertex to find the best value and explain what it means in context. This is especially useful for problems involving area, height, and cost.

Non-Linear Systems of Equations are systems in which at least one equation is not linear. In Grade 11, the most common type is a system with one linear equation and one quadratic equation. Solving the system means finding all ordered pairs \,\((x,y)\)\, that make both equations true at the same time.

These solutions are important because they represent the points of intersection of the graphs. A line and a parabola may intersect in two points, one point, or not at all. Algebra helps us find the exact coordinates, and graphs help us understand what those solutions mean.

In this lesson, you will learn how to solve non-linear systems graphically and algebraically, and how to interpret the answers.

1. What makes a system non-linear?

A linear equation has variables only to the first power, such as \,\(y=2x+1\). Its graph is a straight line.

A quadratic equation includes a squared variable, such as \,\(y=x^2-4\). Its graph is a parabola.

When a system contains one of each, like

$$ \begin{cases} y=x^2-4\\ y=2x+1 \end{cases} $$

the system is non-linear because one equation is quadratic.

2. What do the solutions represent?

Each solution to the system is a point where the graphs meet. That means the same \,\(x\)\, and \,\(y\)\, values satisfy both equations.

• Two solutions: the line crosses the parabola at two points.
• One solution: the line touches the parabola at one point.
• No real solution: the line and parabola do not meet.

This is similar to solving two linear equations, except now the quadratic can cause more than one intersection.

3. Solving graphically

To solve graphically:

1. Graph both equations on the same coordinate plane.
2. Find the intersection point or points.
3. Write the solutions as ordered pairs.

Graphing is useful for estimating solutions and understanding how many solutions there are. However, unless the points are very clear, graphing often gives only approximate answers.

For exact answers, algebraic methods are better.

4. Solving algebraically

The main algebraic idea is to use substitution. Since both equations describe \,\(y\), we can set them equal to each other.

For example, if

$$ \begin{cases} y=x^2-4\\ y=2x+1 \end{cases} $$

then both expressions equal \,\(y\), so:

$$ x^2-4=2x+1 $$

Now solve the quadratic equation. Once you find the \,\(x\)-values, substitute each one back into either original equation to find the matching \,\(y\)-values.

This process usually follows these steps:

1. Rewrite one or both equations so a variable is isolated, if needed.
2. Substitute one equation into the other.
3. Solve the resulting quadratic equation.
4. Substitute each solution back to find the other variable.
5. Check your ordered pairs in both equations.

Worked Example 1: Two intersections

Solve the system:

$$ \begin{cases} y=x^2-1\\ y=x+1 \end{cases} $$

Step 1: Set the equations equal.

$$ x^2-1=x+1 $$

Step 2: Move all terms to one side.

$$ x^2-x-2=0 $$

Step 3: Factor.

$$ (x-2)(x+1)=0 $$

So the \,\(x\)-values are:

$$ x=2 \quad \text{or} \quad x=-1 $$

Step 4: Find the matching \,\(y\)-values. Use \,\(y=x+1\).

If \,\(x=2\), then

$$ y=2+1=3 $$

If \,\(x=-1\), then

$$ y=-1+1=0 $$

Solution:

$$ (2,3) \quad \text{and} \quad (-1,0) $$

This means the line and parabola intersect at two points.

Worked Example 2: One intersection

Solve the system:

$$ \begin{cases} y=x^2\\ y=2x-1 \end{cases} $$

Step 1: Set them equal.

$$ x^2=2x-1 $$

Step 2: Rearrange.

$$ x^2-2x+1=0 $$

Step 3: Factor.

$$ (x-1)^2=0 $$

So there is only one \,\(x\)-value:

$$ x=1 $$

Step 4: Find \,\(y\).

$$ y=2(1)-1=1 $$

Solution:

$$ (1,1) $$

There is only one intersection because the line touches the parabola at exactly one point.

Worked Example 3: No real intersection

Solve the system:

$$ \begin{cases} y=x^2+4\\ y=1 \end{cases} $$

Step 1: Set them equal.

$$ x^2+4=1 $$

Step 2: Solve.

$$ x^2=-3 $$

There is no real number whose square is \,\(-3\). That means there are no real solutions.

Solution: no real solution

Graphically, the parabola \,\(y=x^2+4\)\, is always at or above \,\(y=4\), so it never meets the horizontal line \,\(y=1\).

5. When one equation is not solved for \,\(y\)

Sometimes the linear equation is written in standard form, such as \,\(x+y=5\). In that case, first solve for one variable, then substitute.

Worked Example 4: Rearranging before substitution

Solve the system:

$$ \begin{cases} y=x^2-3x\\ x+y=2 \end{cases} $$

Step 1: Solve the linear equation for \,\(y\).

$$ x+y=2 $$ $$ y=2-x $$

Step 2: Substitute into the quadratic equation.

$$ 2-x=x^2-3x $$

Step 3: Move all terms to one side.

$$ 0=x^2-2x-2 $$

or

$$ x^2-2x-2=0 $$

Step 4: Solve using the quadratic formula.

$$ x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-2)}}{2(1)} $$ $$ x=\frac{2\pm\sqrt{4+8}}{2} $$ $$ x=\frac{2\pm\sqrt{12}}{2}=\frac{2\pm 2\sqrt{3}}{2}=1\pm\sqrt{3} $$

Step 5: Find \,\(y\) using \,\(y=2-x\).

If \,\(x=1+\sqrt{3}\), then

$$ y=2-(1+\sqrt{3})=1-\sqrt{3} $$

If \,\(x=1-\sqrt{3}\), then

$$ y=2-(1-\sqrt{3})=1+\sqrt{3} $$

Solutions:

$$ (1+\sqrt{3},\,1-\sqrt{3}) \quad \text{and} \quad (1-\sqrt{3},\,1+\sqrt{3}) $$

This example shows that some systems do not factor nicely, so the quadratic formula is useful.

6. How to interpret the answers

After solving, always think about what the solutions mean on the graph.

• If you get two ordered pairs, the line crosses the parabola twice.
• If you get one ordered pair, the line just touches the parabola.
• If you get no real solution, the graphs do not intersect.

In applied problems, the solutions may represent times and positions, distances and heights, or other quantities that satisfy two conditions at once.

7. Common mistakes to avoid

• Forgetting to set the equations equal. If both equations equal \,\(y\), then the expressions must equal each other.
• Not moving all terms to one side. You need a quadratic equation in standard form to factor or use the quadratic formula.
• Losing a solution. Quadratics can have two solutions, so check both.
• Using only the \,\(x\)-values. The final answer to a system must be ordered pairs \,\((x,y)\).
• Arithmetic sign errors. Small mistakes with negatives can change the entire result.

8. Quick strategy guide

• If both equations are solved for \,\(y\), set them equal.
• If one equation is not solved for a variable, rearrange it first.
• Solve the quadratic by factoring if possible.
• If factoring does not work, use the quadratic formula.
• Substitute back to find the second coordinate.
• Check your final answers in both equations.

9. Why graphing and algebra work together

Graphing gives a visual picture of the system. You can see whether there should be zero, one, or two intersections. This is helpful for checking if your algebraic answer makes sense.

Algebra gives the exact values of the intersection points. Together, these methods help you understand both the process and the meaning of the solution.

Summary

A non-linear system of equations often includes a line and a parabola. The solutions are the points where the graphs intersect. To solve algebraically, rewrite equations if needed, substitute, solve the resulting quadratic, and then find the corresponding \,\(y\)-values. Depending on the system, there may be two solutions, one solution, or no real solution.