Quadratic Equations

Standard Form of a Quadratic Equation

A quadratic equation is an equation that includes a variable squared, such as \(x^2\). In 10th Grade maths, a quadratic equation is usually written in a special arrangement called standard form.

The standard form of a quadratic equation is:

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

Here, \(a\), \(b\), and \(c\) are numbers, and \(a \ne 0\).

This condition is important. If \(a = 0\), then the \(x^2\) term disappears, and the equation is no longer quadratic.

For example:

• \(2x^2 + 5x - 3 = 0\) is quadratic because the coefficient of \(x^2\) is \(2\), and \(2 \ne 0\).
• \(0x^2 + 5x - 3 = 0\) is not quadratic because the \(x^2\) term is really gone.

Why standard form matters

Writing a quadratic equation in standard form helps you:

• recognize that the equation is quadratic,
• identify the values of \(a\), \(b\), and \(c\),
• use methods like factoring, completing the square, or the quadratic formula later.

The parts of standard form

In

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

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

Example: In \(3x^2 - 7x + 4 = 0\):

• \(a = 3\)
• \(b = -7\)
• \(c = 4\)

Be careful with signs. If a term is negative, its coefficient is negative.

How to rewrite an equation in standard form

Not every quadratic equation starts in standard form. You may need to rearrange it.

To put an equation into standard form:

1. Move all terms to one side so the other side is \(0\).
2. Combine like terms if needed.
3. Write the terms in order: \(x^2\) term, \(x\) term, constant.
4. Check that the coefficient of \(x^2\) is not zero.

Worked Example 1: Already in standard form

Determine whether \(5x^2 + 2x - 8 = 0\) is in standard form, and identify \(a\), \(b\), and \(c\).

Step 1: Check the arrangement.

The equation is written as:

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

This matches \(ax^2 + bx + c = 0\).

Step 2: Identify the coefficients.

• \(a = 5\)
• \(b = 2\)
• \(c = -8\)

Answer: Yes, it is in standard form.

Worked Example 2: Rearranging to standard form

Write \(x^2 + 9 = 4x\) in standard form.

Step 1: Move all terms to one side.

Subtract \(4x\) from both sides:

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

Step 2: Check the order of terms.

The equation is now written in the order \(x^2\), \(x\), constant.

Answer:

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

Here, \(a = 1\), \(b = -4\), and \(c = 9\).

Worked Example 3: Combining like terms

Write \(3x - 2x^2 = 7\) in standard form.

Step 1: Move all terms to one side.

Add \(2x^2\) to both sides and subtract \(7\) from both sides, or more simply rearrange:

$$-2x^2 + 3x - 7 = 0$$

Step 2: Check the order.

The terms are in the correct order: \(x^2\), \(x\), constant.

Answer:

$$-2x^2 + 3x - 7 = 0$$

So:

• \(a = -2\)
• \(b = 3\)
• \(c = -7\)

Notice that \(a\) can be negative. The only rule is that it cannot be zero.

Worked Example 4: More than one side has terms

Write \(4x^2 + 6 = 2x + 1\) in standard form.

Step 1: Move all terms to the left side.

Subtract \(2x\) from both sides and subtract \(1\) from both sides:

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

Step 2: Identify the coefficients.

• \(a = 4\)
• \(b = -2\)
• \(c = 5\)

Answer:

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

Important reminders

• A quadratic equation must have an \(x^2\) term.
• In standard form, one side must be \(0\).
• The terms should be arranged from highest power to lowest power.
• The coefficient \(a\) must not be zero.
• If a term is missing, its coefficient is \(0\). For example, in \(x^2 + 9 = 0\), \(b = 0\).

Example with a missing term

Consider:

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

This is still in standard form because it can be seen as:

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

So:

• \(a = 2\)
• \(b = 0\)
• \(c = -5\)

Common mistakes to avoid

• Forgetting to set the equation equal to 0.
  For example, \(x^2 + 3x = 4\) is not yet in standard form.
• Missing negative signs.
  In \(x^2 - 6x + 2 = 0\), the value of \(b\) is \(-6\), not \(6\).
• Thinking \(a\) must be positive.
  It does not have to be positive. It only must not be zero.
• Not writing terms in order.
  For example, \(3 - 2x + x^2 = 0\) is equivalent, but standard form is better written as \(x^2 - 2x + 3 = 0\).

Quick check

Decide whether each equation is in standard form:

• \(x^2 + 7x - 1 = 0\) → Yes
• \(2x + x^2 = 5\) → No
• \(-3x^2 + 4 = 0\) → Yes
• \(6x - 9 = 0\) → No, because there is no \(x^2\) term

Summary

The standard form of a quadratic equation is:

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

To write an equation in this form, move all terms to one side, combine like terms, and arrange the terms in descending order. Always check that \(a \ne 0\), because that is what makes the equation quadratic.

Solving Quadratics by Factorization

A quadratic equation is an equation whose highest power of the variable is 2. A common form is:

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

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

One important way to solve quadratic equations is factorization. This means rewriting the quadratic expression as a product of two simpler expressions.

For example, instead of working with \(x^2 + 5x + 6\), we can rewrite it as:

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

Once a quadratic is written as a product, we can use the zero-product property.

Zero-Product Property: If two numbers multiply to give 0, then at least one of them must be 0.

In algebra, this means:

$$ab = 0 \quad \Rightarrow \quad a=0 \text{ or } b=0$$

This idea is the key to solving quadratics by factorization.

Why factorization works

If we have an equation like:

$$ (x+2)(x+3)=0 $$

then either:

• \(x+2=0\), which gives \(x=-2\),
• or \(x+3=0\), which gives \(x=-3\).

So the solutions to the quadratic are \(-2\) and \(-3\).

General steps for solving by factorization

1. Write the equation in standard form: $$ax^2 + bx + c = 0$$
2. Factor the quadratic expression.
3. Apply the zero-product property.
4. Solve each linear equation.
5. Check your answers by substituting them back into the original equation.

How to factor simple quadratics

For a quadratic like \(x^2+bx+c\), we look for two numbers that:

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

Example: To factor \(x^2+7x+12\), we need two numbers that multiply to 12 and add to 7.

Those numbers are 3 and 4, so:

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

Worked Example 1: A basic quadratic

Solve:

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

Step 1: Factor the quadratic.

We need two numbers that multiply to 6 and add to 5. These are 2 and 3.

So:

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

Step 2: Use the zero-product property.

$$ (x+2)(x+3)=0 $$

This means:

$$x+2=0 \quad \text{or} \quad x+3=0$$

Step 3: Solve each equation.

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

Answer: The solutions are \(x=-2\) and \(x=-3\).

Check:

Substitute \(x=-2\):

$$(-2)^2 + 5(-2) + 6 = 4 -10 + 6 = 0$$

Substitute \(x=-3\):

$$(-3)^2 + 5(-3) + 6 = 9 -15 + 6 = 0$$

Both values work.

Worked Example 2: When the coefficient of \(x^2\) is 1 but signs are different

Solve:

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

Step 1: Factor.

We need two numbers that multiply to \(-12\) and add to \(-1\).

These numbers are \(-4\) and \(3\).

So:

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

Step 2: Apply the zero-product property.

$$ (x-4)(x+3)=0 $$

So:

$$x-4=0 \quad \text{or} \quad x+3=0$$

Step 3: Solve.

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

Answer: \(x=4\) or \(x=-3\).

Worked Example 3: When the coefficient of \(x^2\) is not 1

Solve:

$$2x^2 + 7x + 3 = 0$$

Now the first term is \(2x^2\), so factoring takes a little more care.

We want two brackets whose first terms multiply to \(2x^2\) and whose last terms multiply to 3.

Try:

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

Check by expanding:

$$ (2x+1)(x+3)=2x^2+6x+x+3=2x^2+7x+3 $$

So the factorization is correct.

Step 2: Apply the zero-product property.

$$ (2x+1)(x+3)=0 $$

So:

$$2x+1=0 \quad \text{or} \quad x+3=0$$

Step 3: Solve each equation.

From \(2x+1=0\):

$$2x=-1$$

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

From \(x+3=0\):

$$x=-3$$

Answer: The solutions are \(x=-\frac{1}{2}\) and \(x=-3\).

Worked Example 4: First move everything to one side

Solve:

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

This equation is not yet in standard form. Before factoring, move everything to one side.

Subtract \(6x\) from both sides:

$$x^2 - 6x + 9 = 0$$

Step 2: Factor.

We need two numbers that multiply to 9 and add to \(-6\). These are \(-3\) and \(-3\).

So:

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

Step 3: Apply the zero-product property.

$$ (x-3)^2=0 $$

This means:

$$x-3=0$$

So:

$$x=3$$

Answer: The equation has one repeated solution, \(x=3\).

Important reminders

• Always make one side equal to 0 before factoring.
• Factor completely. Do not stop halfway.
• After factoring, set each factor equal to 0.
• Be careful with positive and negative signs.
• Check your solutions in the original equation when possible.

Common mistakes to avoid

• Forgetting to write the equation as \(=0\) before factoring.
• Only solving one factor and missing the other solution.
• Mixing up the signs when choosing factor pairs.
• Incorrect factoring, especially when the coefficient of \(x^2\) is not 1.

Quick factor-pair thinking

When factoring, it helps to list pairs of numbers.

For example, if you need numbers that multiply to 12, the possible pairs are:

• 1 and 12
• 2 and 6
• 3 and 4

Then choose the pair that gives the correct middle term when added or subtracted.

When factorization works best

Factorization is very useful when the quadratic can be factored into simple linear expressions. Not every quadratic factors easily using whole numbers, but many school problems are designed so that they do.

If a quadratic does factor, solving by factorization is often faster than other methods.

Summary

To solve a quadratic by factorization, first write it in the form:

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

Next, factor the quadratic into two brackets. Then use the zero-product property: if the product is 0, at least one factor must be 0. Solve each resulting linear equation to get the roots.

With practice, you will get faster at spotting factor pairs and solving quadratics accurately.

Solving Quadratics by Completing the Square

A quadratic equation is an equation that can be written in the form \(ax^2+bx+c=0\), where \(a\neq 0\). One way to solve quadratic equations is called completing the square.

This method changes a quadratic expression into a perfect square trinomial, which is an expression that factors neatly into something like \((x+3)^2\) or \((x-5)^2\).

Completing the square is important because it helps you solve equations, understand where the quadratic formula comes from, and rewrite quadratics in a form that is often easier to graph.

1. What is a perfect square trinomial?

A perfect square trinomial has one of these patterns:

• \(x^2+2ax+a^2=(x+a)^2\)
• \(x^2-2ax+a^2=(x-a)^2\)

For example:

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

The key idea is that if you have \(x^2+bx\), you can turn it into a perfect square by adding a special number.

That special number is:

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

So for \(x^2+10x\), the number to add is:

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

This gives:

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

2. The steps for completing the square

To solve a quadratic equation by completing the square, follow these steps.

1. Make sure the equation is in the form \(ax^2+bx+c=0\).
2. If \(a\neq 1\), divide every term by \(a\).
3. Move the constant term to the other side.
4. Take half of the coefficient of \(x\), then square it.
5. Add that number to both sides of the equation.
6. Rewrite the left side as a squared binomial.
7. Take the square root of both sides.
8. Solve for \(x\).

Why do we add the same number to both sides?

Because an equation must stay balanced. If you add a number to only one side, the equation changes. Adding the same number to both sides keeps the equation equivalent.

3. Worked Example 1: A simple quadratic

Solve:

$$x^2+6x-7=0$$

Step 1: Move the constant term to the other side.

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

Step 2: Take half of 6 and square it.

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

Step 3: Add 9 to both sides.

$$x^2+6x+9=7+9$$ $$x^2+6x+9=16$$

Step 4: Rewrite the left side as a perfect square.

$$ (x+3)^2=16 $$

Step 5: Take the square root of both sides.

$$x+3=\pm 4$$

Step 6: Solve for \(x\).

• \(x+3=4 \Rightarrow x=1\)
• \(x+3=-4 \Rightarrow x=-7\)

Answer: \(x=1\) or \(x=-7\)

4. Worked Example 2: A quadratic with a negative middle term

Solve:

$$x^2-10x+9=0$$

Step 1: Move the constant term.

$$x^2-10x=-9$$

Step 2: Take half of \(-10\) and square it.

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

Step 3: Add 25 to both sides.

$$x^2-10x+25=-9+25$$ $$x^2-10x+25=16$$

Step 4: Rewrite.

$$ (x-5)^2=16 $$

Step 5: Take square roots.

$$x-5=\pm 4$$

Step 6: Solve.

• \(x-5=4 \Rightarrow x=9\)
• \(x-5=-4 \Rightarrow x=1\)

Answer: \(x=9\) or \(x=1\)

5. Worked Example 3: When the coefficient of \(x^2\) is not 1

Solve:

$$2x^2+8x-10=0$$

Before completing the square, the coefficient of \(x^2\) should be 1.

Step 1: Divide every term by 2.

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

Step 2: Move the constant term.

$$x^2+4x=5$$

Step 3: Take half of 4 and square it.

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

Step 4: Add 4 to both sides.

$$x^2+4x+4=5+4$$ $$x^2+4x+4=9$$

Step 5: Rewrite.

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

Step 6: Take square roots.

$$x+2=\pm 3$$

Step 7: Solve.

• \(x+2=3 \Rightarrow x=1\)
• \(x+2=-3 \Rightarrow x=-5\)

Answer: \(x=1\) or \(x=-5\)

6. Worked Example 4: Solutions that are not integers

Solve:

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

Step 1: Move the constant term.

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

Step 2: Take half of 2 and square it.

$$\left(\frac{2}{2}\right)^2=1^2=1$$

Step 3: Add 1 to both sides.

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

Step 4: Rewrite.

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

Step 5: Take square roots.

$$x+1=\pm \sqrt{2}$$

Step 6: Solve.

$$x=-1\pm \sqrt{2}$$

Answer: \(x=-1+\sqrt{2}\) or \(x=-1-\sqrt{2}\)

7. Common mistakes to avoid

• Forgetting to move the constant first: It is easier to complete the square when the \(x^2\) and \(x\) terms are on one side.
• Not dividing by the coefficient of \(x^2\): If the first coefficient is not 1, divide first.
• Adding the square term to only one side: Always add it to both sides.
• Forgetting the \(\pm\) when taking square roots: If \(x^2=16\), then \(x=4\) and \(x=-4\).
• Squaring the wrong value: You must take half of the coefficient of \(x\), then square it.

8. Quick check: turning an expression into a perfect square

Find the number that makes each expression a perfect square trinomial.

• \(x^2+8x+\square\)
• Half of 8 is 4, and \(4^2=16\)
• So: \(x^2+8x+16=(x+4)^2\)

• \(x^2-12x+\square\)
• Half of \(-12\) is \(-6\), and \((-6)^2=36\)
• So: \(x^2-12x+36=(x-6)^2\)

9. Why this method works

Completing the square works because it changes a quadratic expression into a form that is easier to solve:

$$x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2$$

Once the left side is a square, you can use square roots to solve the equation.

10. Final summary

To solve a quadratic by completing the square, first make the coefficient of \(x^2\) equal to 1. Then move the constant term, add \(\left(\frac{b}{2}\right)^2\) to both sides, rewrite the trinomial as a square, and solve using square roots.

This method is especially helpful when factoring is difficult. The most important ideas are to keep the equation balanced, use the correct number when completing the square, and remember the \(\pm\) after taking square roots.

Derivation and Use of the Quadratic Formula

A quadratic equation is an equation that can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are numbers and \(a \ne 0\).

Some quadratic equations can be solved by factoring, but not all of them factor easily. That is why the quadratic formula is so useful. It gives a method that works for every quadratic equation.

In this lesson, you will learn:

• where the quadratic formula comes from,
• how to derive it by completing the square,
• how to use it correctly, and
• how the expression inside the square root tells us about the roots.

1. Start with the standard form

To use or derive the quadratic formula, the equation must first be written in standard form:

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

Here, \(a\), \(b\), and \(c\) are constants, and \(a \ne 0\).

2. Deriving the quadratic formula

We begin with the general quadratic equation:

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

The goal is to solve for \(x\) by completing the square.

Step 1: Move the constant term to the other side.

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

Step 2: Divide every term by \(a\) so the coefficient of \(x^2\) becomes 1.

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

Step 3: Complete the square.

Take half of the coefficient of \(x\), then square it:

$$\left(\frac{1}{2} \cdot \frac{b}{a}\right)^2 = \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$$

Add this value 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:

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

Rewrite the right side with a common denominator of \(4a^2\):

$$-\frac{c}{a} = -\frac{4ac}{4a^2}$$

So:

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

Step 4: Take the square root of both sides.

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

Step 5: Solve for \(x\).

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

Combine the terms over a common denominator:

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

This is the quadratic formula.

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

3. What each part means

• \(-b\) is the opposite of the \(x\)-coefficient
• \(2a\) is twice the coefficient of \(x^2\)
• \(b^2 - 4ac\) is called the discriminant
• The symbol \(\pm\) means there may be two solutions: one using \(+\) and one using \(-\)

4. The discriminant and the nature of the roots

The discriminant is:

$$b^2 - 4ac$$

It tells us how many real solutions the quadratic equation has.

• If \(b^2 - 4ac > 0\), there are two different real solutions.
• If \(b^2 - 4ac = 0\), there is one real repeated solution.
• If \(b^2 - 4ac < 0\), there are no real solutions.

This is helpful because even before solving, you can tell what kind of answer to expect.

5. How to use the quadratic formula

Follow these steps each time:

1. Write the equation in standard form: \(ax^2 + bx + c = 0\)
2. Identify \(a\), \(b\), and \(c\)
3. Substitute them into the formula
4. Simplify carefully
5. Check if the square root can be simplified
6. Write both solutions if there are two

Important note: Be very careful with negative signs, especially for \(b\). If \(b = -5\), then \(-b = -(-5) = 5\).

Worked Example 1: A simple quadratic

Solve:

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

Here:

$$a = 1, \quad b = 5, \quad c = 6$$

Use the formula:

$$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$$ $$x = \frac{-5 \pm \sqrt{25 - 24}}{2}$$ $$x = \frac{-5 \pm \sqrt{1}}{2}$$ $$x = \frac{-5 \pm 1}{2}$$

Now find both values:

$$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$$ $$x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$$

Answer: \(x = -2\) or \(x = -3\)

Even though this quadratic could be factored, the quadratic formula still works.

Worked Example 2: A quadratic that does not factor easily

Solve:

$$2x^2 + 3x - 7 = 0$$

Identify the values:

$$a = 2, \quad b = 3, \quad c = -7$$

Substitute into the formula:

$$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)}$$ $$x = \frac{-3 \pm \sqrt{9 + 56}}{4}$$ $$x = \frac{-3 \pm \sqrt{65}}{4}$$

Since \(65\) is not a perfect square, this is the exact form.

Answer:

$$x = \frac{-3 + \sqrt{65}}{4} \quad \text{or} \quad x = \frac{-3 - \sqrt{65}}{4}$$

This example shows why the quadratic formula is important. Factoring would be difficult, but the formula gives the solutions directly.

Worked Example 3: One repeated solution

Solve:

$$x^2 - 6x + 9 = 0$$

Identify the values:

$$a = 1, \quad b = -6, \quad c = 9$$

Substitute into the formula:

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(9)}}{2(1)}$$ $$x = \frac{6 \pm \sqrt{36 - 36}}{2}$$ $$x = \frac{6 \pm \sqrt{0}}{2}$$ $$x = \frac{6}{2} = 3$$

Answer: \(x = 3\)

Because the discriminant is 0, there is only one real repeated root.

Worked Example 4: No real solutions

Solve:

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

Identify the values:

$$a = 1, \quad b = 4, \quad c = 8$$

Substitute into the formula:

$$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(8)}}{2(1)}$$ $$x = \frac{-4 \pm \sqrt{16 - 32}}{2}$$ $$x = \frac{-4 \pm \sqrt{-16}}{2}$$

The number under the square root is negative. That means there are no real solutions.

Answer: No real solutions

6. Common mistakes to avoid

• Forgetting to write the equation in standard form. Always make sure one side is 0.
• Using the wrong values for \(a\), \(b\), or \(c\). Pay attention to signs.
• Forgetting parentheses when substituting. For example, write \((-6)^2\), not \(-6^2\).
• Missing the \(\pm\). This can make you lose one of the solutions.
• Not simplifying the square root correctly. Leave answers exact unless you are told to round.

7. When should you use the quadratic formula?

You should especially use the quadratic formula when:

• the equation does not factor easily,
• you want an exact answer,
• you want to know the number of real solutions using the discriminant.

It is one of the most reliable ways to solve a quadratic equation.

8. Quick check: identify the discriminant

For each equation, look at \(b^2 - 4ac\).

• For \(x^2 - 2x - 3 = 0\): \(( -2)^2 - 4(1)(-3) = 4 + 12 = 16\) → two real solutions
• For \(x^2 + 10x + 25 = 0\): \(10^2 - 4(1)(25) = 100 - 100 = 0\) → one repeated real solution
• For \(x^2 + 2x + 5 = 0\): \(2^2 - 4(1)(5) = 4 - 20 = -16\) → no real solutions

9. Summary

The quadratic formula is derived by completing the square on the general equation \(ax^2 + bx + c = 0\). The result is:

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

This formula can solve any quadratic equation, even when factoring is difficult or impossible.

The discriminant, \(b^2 - 4ac\), tells you whether the equation has two real solutions, one repeated real solution, or no real solutions.

If you carefully identify \(a\), \(b\), and \(c\), substitute with the correct signs, and simplify step by step, you can solve quadratic equations confidently using the quadratic formula.

Lesson: The Discriminant and Nature of Roots

When we solve a quadratic equation, we are trying to find the values of the variable that make the equation true. A quadratic equation is usually written in the form:

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

Here, a, b, and c are numbers, and a  cannot be 0.

Some quadratic equations have two different real solutions, some have one repeated real solution, and some have no real solutions. The discriminant helps us tell which kind of roots a quadratic has without fully solving it.

The discriminant comes from the quadratic formula:

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

The expression inside the square root, \(b^2 - 4ac\), is called the discriminant.

We usually give the discriminant the symbol \(D\), so:

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

The value of the discriminant tells us about the nature of the roots, which means the type of solutions the quadratic equation has.

Main idea:

• If \(D > 0\), the equation has two real and distinct roots.
• If \(D = 0\), the equation has two real and equal roots (also called one repeated root).
• If \(D < 0\), the equation has two complex and non-real roots.

This happens because of the square root in the quadratic formula.

• A positive number under the square root gives a real answer.
• A zero under the square root gives 0, so both roots become the same.
• A negative number under the square root does not give a real number.

So, by calculating just one value, \(b^2 - 4ac\), we can quickly classify the roots.

How to use the discriminant

1. Write the equation in the form \(ax^2 + bx + c = 0\).
2. Identify the values of \(a\), \(b\), and \(c\).
3. Substitute them into \(D = b^2 - 4ac\).
4. Decide whether \(D\) is positive, zero, or negative.
5. Use that to describe the roots.

Important note: Be careful with signs. If \(b\) or \(c\) is negative, include the negative sign when substituting.

Worked Example 1: Two real and distinct roots

Find the nature of the roots of:

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

First, identify the values:

• \(a = 1\)
• \(b = -5\)
• \(c = 6\)

Now calculate the discriminant:

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

$$D = (-5)^2 - 4(1)(6)$$

$$D = 25 - 24 = 1$$

Since \(D = 1\), and 1 is greater than 0, the equation has two real and distinct roots.

Worked Example 2: Real and equal roots

Find the nature of the roots of:

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

Identify the values:

• \(a = 1\)
• \(b = -4\)
• \(c = 4\)

Calculate the discriminant:

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

$$D = (-4)^2 - 4(1)(4)$$

$$D = 16 - 16 = 0$$

Since \(D = 0\), the equation has real and equal roots.

This means the quadratic touches the x-axis at exactly one point, and both solutions are the same number.

Worked Example 3: Complex and non-real roots

Find the nature of the roots of:

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

Identify the values:

• \(a = 2\)
• \(b = 3\)
• \(c = 5\)

Calculate the discriminant:

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

$$D = 3^2 - 4(2)(5)$$

$$D = 9 - 40 = -31$$

Since \(D = -31\), and this is less than 0, the equation has two complex and non-real roots.

This means there are no real values of \(x\) that make the equation equal to 0.

Worked Example 4: Be careful with negative signs

Find the nature of the roots of:

$$3x^2 + 2x - 1 = 0$$

Identify the values:

• \(a = 3\)
• \(b = 2\)
• \(c = -1\)

Now substitute carefully:

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

$$D = 2^2 - 4(3)(-1)$$

$$D = 4 + 12 = 16$$

Since \(D = 16 > 0\), the equation has two real and distinct roots.

Notice that subtracting a negative number became addition. This is a very common place for mistakes, so always check the signs carefully.

What the discriminant tells us about the graph

A quadratic equation is connected to the graph of a parabola. The roots are the x-values where the graph crosses or touches 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 is another useful way to understand the nature of roots.

Common mistakes to avoid

• Not writing the equation in standard form first. For example, move all terms to one side before choosing \(a\), \(b\), and \(c\).
• Forgetting that \(b^2\) means the whole value of \(b\) is squared. For example, if \(b = -6\), then \(b^2 = (-6)^2 = 36\), not \(-36\).
• Sign errors with \(-4ac\). Always use brackets when substituting.
• Confusing “equal roots” with “no roots.” If the discriminant is 0, the roots are real and equal, not missing.

Quick check table

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

• \(D > 0\)  two real and distinct roots
• \(D = 0\)  two real and equal roots
• \(D < 0\)  two complex and non-real roots

Practice-style questions to think about

Try classifying the roots of these equations by finding the discriminant:

1. \(x^2 + 6x + 5 = 0\)
2. \(x^2 + 8x + 16 = 0\)
3. \(x^2 + 2x + 10 = 0\)

You can check your thinking like this:

• For \(x^2 + 6x + 5 = 0\): \(D = 36 - 20 = 16 > 0\), so two real and distinct roots.
• For \(x^2 + 8x + 16 = 0\): \(D = 64 - 64 = 0\), so real and equal roots.
• For \(x^2 + 2x + 10 = 0\): \(D = 4 - 40 = -36 < 0\), so complex and non-real roots.

Summary

The discriminant is the part of the quadratic formula given by \(b^2 - 4ac\). It tells us the type of roots a quadratic equation has without solving the whole equation.

Remember:

• Positive discriminant  two real and distinct roots
• Zero discriminant  real and equal roots
• Negative discriminant  complex and non-real roots

If you can identify \(a\), \(b\), and \(c\) correctly and substitute carefully, you can use the discriminant confidently to describe the nature of roots in any quadratic equation.

Forming Quadratic Equations from Roots

Sometimes, instead of solving a quadratic equation to find its roots, we do the reverse: we are given the roots and asked to form the quadratic equation.

This idea is very useful because roots and factors are closely connected. If we know the roots, we can build the factors, and from the factors, we can write the quadratic equation.

In this lesson, you will learn:

• how roots and factors are related,
• how to form a quadratic equation from given roots,
• how to use the sum and product of roots,
• how to handle whole number, negative, and fractional roots.

1. The key idea: roots and factors

If a number \(r\) is a root of a quadratic equation, then \((x-r)\) is a factor of that equation.

So:

• if \(r\) is a root, one factor is \((x-r)\),
• if \(s\) is another root, another factor is \((x-s)\).

That means a quadratic equation with roots \(r\) and \(s\) can be written as:

$$ (x-r)(x-s)=0 $$

When we expand this, we get the quadratic equation.

2. Expanding the factors

Let the roots be \(\alpha\) and \(\beta\). Then:

$$ (x-\alpha)(x-\beta)=0 $$

Expanding:

$$ x^2-(\alpha+\beta)x+\alpha\beta=0 $$

This gives us a very important result:

• Sum of roots = \(\alpha+\beta\)
• Product of roots = \(\alpha\beta\)

So the quadratic equation formed from roots \(\alpha\) and \(\beta\) is:

$$ x^2-(\text{sum of roots})x+(\text{product of roots})=0 $$

3. Main formula

If the roots are \(\alpha\) and \(\beta\), then the quadratic equation is:

$$ x^2-(\alpha+\beta)x+\alpha\beta=0 $$

This is the quickest method when the leading coefficient is \(1\).

4. Step-by-step method

To form a quadratic equation from its roots:

1. Write the two roots.
2. Find their sum.
3. Find their product.
4. Substitute into:

$$ x^2-(\text{sum})x+(\text{product})=0 $$

You can also form the equation by writing the factors first:

$$ (x-\alpha)(x-\beta)=0 $$

Then expand and simplify.

5. Worked Examples

Example 1: Roots are 2 and 5

We are given the roots:

$$ \alpha=2, \quad \beta=5 $$

Find the sum of the roots:

$$ \alpha+\beta=2+5=7 $$

Find the product of the roots:

$$ \alpha\beta=2\times 5=10 $$

Substitute into the formula:

$$ x^2-(\alpha+\beta)x+\alpha\beta=0 $$ $$ x^2-7x+10=0 $$

Answer: The required quadratic equation is:

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

Let us check using factors:

$$ (x-2)(x-5)=0 $$ $$ x^2-5x-2x+10=0 $$ $$ x^2-7x+10=0 $$

The result matches.

Example 2: Roots are -3 and 4

Now one root is negative.

$$ \alpha=-3, \quad \beta=4 $$

Find the sum:

$$ \alpha+\beta=-3+4=1 $$

Find the product:

$$ \alpha\beta=(-3)(4)=-12 $$

Use the formula:

$$ x^2-(1)x+(-12)=0 $$ $$ x^2-x-12=0 $$

Answer:

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

Check with factors:

$$ (x-(-3))(x-4)=0 $$

Since \(x-(-3)=x+3\), we get:

$$ (x+3)(x-4)=0 $$ $$ x^2-4x+3x-12=0 $$ $$ x^2-x-12=0 $$

This confirms the equation.

Example 3: Roots are \(\frac{1}{2}\) and 3

Fractional roots are handled in the same way.

$$ \alpha=\frac{1}{2}, \quad \beta=3 $$

Find the sum:

$$ \alpha+\beta=\frac{1}{2}+3=\frac{1}{2}+\frac{6}{2}=\frac{7}{2} $$

Find the product:

$$ \alpha\beta=\frac{1}{2}\times 3=\frac{3}{2} $$

Substitute into the formula:

$$ x^2-\frac{7}{2}x+\frac{3}{2}=0 $$

This is correct, but we usually prefer an equation with integer coefficients. Multiply every term by 2:

$$ 2\left(x^2-\frac{7}{2}x+\frac{3}{2}\right)=0 $$ $$ 2x^2-7x+3=0 $$

Answer:

$$ 2x^2-7x+3=0 $$

Check with factors:

$$ \left(x-\frac{1}{2}\right)(x-3)=0 $$

To avoid fractions in factors, write \(x-\frac{1}{2}\) as \(2x-1\) when forming an equivalent equation:

$$ (2x-1)(x-3)=0 $$ $$ 2x^2-6x-x+3=0 $$ $$ 2x^2-7x+3=0 $$

Example 4: Roots are 6 and 6

Here both roots are the same. This is called an equal root or repeated root.

$$ \alpha=6, \quad \beta=6 $$

Find the sum:

$$ \alpha+\beta=6+6=12 $$

Find the product:

$$ \alpha\beta=6\times 6=36 $$

Use the formula:

$$ x^2-12x+36=0 $$

Answer:

$$ x^2-12x+36=0 $$

This can also be written from factors as:

$$ (x-6)(x-6)=0 $$ $$ (x-6)^2=0 $$

6. Important sign reminder

Be very careful with signs.

• If the root is \(4\), the factor is \((x-4)\).
• If the root is \(-4\), the factor is \((x+4)\).

This is a common place where students make mistakes.

For example, if the roots are \(-2\) and \(-7\), then:

$$ (x+2)(x+7)=0 $$

Expanding:

$$ x^2+9x+14=0 $$

7. Forming the equation directly from sum and product

Sometimes the question gives the sum and product of the roots instead of the roots themselves.

If:

• sum of roots = \(S\)
• product of roots = \(P\)

then the quadratic equation is:

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

For example, if the sum is \(8\) and product is \(15\), then:

$$ x^2-8x+15=0 $$

8. Quick comparison of the two methods

You can form a quadratic equation in either of these ways:

• Factor method: Write \((x-r)(x-s)=0\), then expand.
• Sum-product method: Use \(x^2-(r+s)x+rs=0\).

Both methods give the same result. The sum-product method is often faster.

9. Common mistakes to avoid

• Wrong sign in the factor: root \(3\) gives factor \((x-3)\), not \((x+3)\).
• Forgetting the minus in the formula: it is \(x^2-(\text{sum})x+(\text{product})=0\).
• Errors with negative numbers: check sums and products carefully.
• Leaving fractions when integer coefficients are expected: multiply through by the denominator.

10. Practice questions

Try these on your own:

1. Form the quadratic equation with roots \(1\) and \(7\).
2. Form the quadratic equation with roots \(-5\) and \(2\).
3. Form the quadratic equation with roots \(\frac{3}{2}\) and \(-1\).
4. Form the quadratic equation whose sum of roots is \(10\) and product is \(21\).

Answers:

1. \(x^2-8x+7=0\)
2. \(x^2+3x-10=0\)
3. \(2x^2-x-3=0\)
4. \(x^2-10x+21=0\)

Summary

To form a quadratic equation from roots \(\alpha\) and \(\beta\), use the factors \((x-\alpha)(x-\beta)\) or the formula:

$$ x^2-(\alpha+\beta)x+\alpha\beta=0 $$

The coefficient of \(x\) comes from the sum of the roots, and the constant term comes from the product of the roots.

If the roots are fractions, you may need to multiply through to remove denominators. Always check signs carefully, especially when roots are negative.

Equations Reducible to Quadratic Form

Sometimes an equation does not look like a quadratic equation at first. It may have powers like \(x^4\), or expressions such as \(x + \frac{1}{x}\), or even fractions with variables. But after a clever substitution, the equation can be turned into a quadratic equation.

These are called equations reducible to quadratic form. The main idea is to spot a repeated pattern, replace it with a temporary variable, solve the quadratic, and then substitute back.

This skill is important because many equations that seem difficult become much easier once you recognize the hidden quadratic structure.

1. Review: What is a quadratic equation?

A quadratic equation is an equation that can be written in the form

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

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

We usually solve quadratic equations by factoring, completing the square, or using the quadratic formula.

2. What does “reducible to quadratic form” mean?

An equation is reducible to quadratic form if we can rewrite it so that it becomes quadratic in some new variable.

For example, suppose we have

$$x^4 - 5x^2 + 6 = 0$$

This is not quadratic in \(x\), because the highest power is \(4\). But it is quadratic in \(x^2\). If we let

$$u = x^2$$

then the equation becomes

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

Now we can solve it like a normal quadratic.

3. When should you use substitution?

Look for equations where powers or expressions repeat in a pattern.

• Expressions like \(x^4\), \(x^2\), and constants
• Expressions like \(x^6\), \(x^3\), and constants
• Expressions involving \(x + \frac{1}{x}\)
• Rational equations that can be rearranged into a repeated form

Common substitutions include:

• \(u = x^2\)
• \(u = x^3\)
• \(u = x + \frac{1}{x}\)
• \(u = \frac{1}{x}\)

4. General method

1. Identify the repeating expression.
2. Let that expression equal a new variable, such as \(u\).
3. Rewrite the equation in terms of \(u\).
4. Solve the quadratic equation in \(u\).
5. Substitute back to find the values of \(x\).
6. Check your answers, especially if the original equation had fractions or square roots.

5. Worked Example 1: A basic hidden quadratic

Solve:

$$x^4 - 5x^2 + 6 = 0$$

Step 1: Choose a substitution.

Because both \(x^4\) and \(x^2\) appear, let

$$u = x^2$$

Then \(x^4 = (x^2)^2 = u^2\). The equation becomes

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

Step 2: Solve the quadratic.

Factor:

$$u^2 - 5u + 6 = (u - 2)(u - 3) = 0$$

So

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

Step 3: Substitute back.

Since \(u = x^2\), we get

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

Therefore,

$$x = \pm \sqrt{2}, \quad x = \pm \sqrt{3}$$

Answer: \(x = \pm\sqrt{2}, \pm\sqrt{3}\)

6. Worked Example 2: Higher powers

Solve:

$$x^6 - 10x^3 + 9 = 0$$

Step 1: Spot the pattern.

The powers are \(x^6\) and \(x^3\). Since \(x^6 = (x^3)^2\), let

$$u = x^3$$

Then the equation becomes

$$u^2 - 10u + 9 = 0$$

Step 2: Solve the quadratic.

$$u^2 - 10u + 9 = (u - 1)(u - 9) = 0$$

So

$$u = 1 \quad \text{or} \quad u = 9$$

Step 3: Substitute back.

Since \(u = x^3\), we get

$$x^3 = 1 \quad \text{or} \quad x^3 = 9$$

Now solve each:

$$x = 1 \quad \text{or} \quad x = \sqrt[3]{9}$$

Answer: \(x = 1\) or \(x = \sqrt[3]{9}\)

7. Worked Example 3: A rational equation

Solve:

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

This equation is not obviously quadratic, but it has a useful pattern.

We know that

$$\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$$

From the given equation,

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

so add \(2\) to both sides:

$$x^2 + 2 + \frac{1}{x^2} = 7$$

Therefore,

$$\left(x + \frac{1}{x}\right)^2 = 7$$

So

$$x + \frac{1}{x} = \pm \sqrt{7}$$

Now solve each case separately.

Case 1:

$$x + \frac{1}{x} = \sqrt{7}$$

Multiply by \(x\):

$$x^2 + 1 = \sqrt{7}x$$ $$x^2 - \sqrt{7}x + 1 = 0$$

Case 2:

$$x + \frac{1}{x} = -\sqrt{7}$$

Multiply by \(x\):

$$x^2 + 1 = -\sqrt{7}x$$ $$x^2 + \sqrt{7}x + 1 = 0$$

Now solve both quadratics.

For \(x^2 - \sqrt{7}x + 1 = 0\):

$$x = \frac{\sqrt{7} \pm \sqrt{7 - 4}}{2} = \frac{\sqrt{7} \pm \sqrt{3}}{2}$$

For \(x^2 + \sqrt{7}x + 1 = 0\):

$$x = \frac{-\sqrt{7} \pm \sqrt{7 - 4}}{2} = \frac{-\sqrt{7} \pm \sqrt{3}}{2}$$

Answer:

$$x = \frac{\sqrt{7} + \sqrt{3}}{2}, \; \frac{\sqrt{7} - \sqrt{3}}{2}, \; \frac{-\sqrt{7} + \sqrt{3}}{2}, \; \frac{-\sqrt{7} - \sqrt{3}}{2}$$

Important: Since the original equation contains \(\frac{1}{x^2}\), \(x \ne 0\). None of these answers is zero, so they are all allowed.

8. Worked Example 4: A fractional equation reducible by substitution

Solve:

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

Step 1: Remove the fraction carefully.

Since \(x \ne 0\), multiply both sides by \(x^2\):

$$2x^4 + 3 = 5x^2$$

Rearrange:

$$2x^4 - 5x^2 + 3 = 0$$

Step 2: Substitute.

Let

$$u = x^2$$

Then:

$$2u^2 - 5u + 3 = 0$$

Step 3: Solve the quadratic.

$$2u^2 - 5u + 3 = (2u - 3)(u - 1) = 0$$

So

$$u = \frac{3}{2} \quad \text{or} \quad u = 1$$

Step 4: Substitute back.

$$x^2 = \frac{3}{2} \quad \text{or} \quad x^2 = 1$$

Therefore,

$$x = \pm \sqrt{\frac{3}{2}}, \quad x = \pm 1$$

Answer: \(x = \pm 1, \pm \sqrt{\frac{3}{2}}\)

9. Things to watch out for

• Do not forget to substitute back. Solving for \(u\) is not the final answer unless the question asks for \(u\).
• Check restrictions. If the original equation has \(\frac{1}{x}\) or \(\frac{1}{x^2}\), then \(x \ne 0\).
• Look for all solutions. If you get \(x^2 = 4\), then \(x = 2\) and \(x = -2\).
• Be careful with powers. If \(u = x^3\), then solving \(u = 8\) gives \(x = 2\), not \(x = \pm 2\).

10. Quick problem-solving checklist

1. Does the equation have a repeated pattern?
2. Can I replace that pattern with a variable like \(u\)?
3. Does the new equation become quadratic?
4. After solving, did I replace \(u\) with the original expression?
5. Did I check for excluded values, such as \(x = 0\)?

11. Summary

Equations reducible to quadratic form are equations that can be turned into a quadratic by using substitution. The key is to recognize a repeated expression, such as \(x^2\), \(x^3\), or another pattern, and replace it with a temporary variable.

Once the equation becomes quadratic, solve it as usual, then substitute back to find the original variable. Always check for restrictions and make sure your final answers satisfy the original equation.

Modeling Word Problems: Geometry and Optimization

In many real-life problems, you are not given a quadratic equation directly. Instead, you are given a situation involving shapes, lengths, area, or maximum and minimum values. Your job is to model the situation with an equation and then solve it.

In this lesson, you will learn how to turn geometry word problems into quadratic equations. You will also learn how to solve optimization problems, where you must find the largest area, the smallest length, or another best value.

These problems often use ideas you already know:

• Area formulas
• Perimeter formulas
• The Pythagorean theorem
• Factoring and the quadratic formula

When geometry and algebra work together, quadratics appear naturally.

1. A Strategy for Modeling Word Problems

When solving a geometry word problem, do not rush to solve right away. First, organize the information.

1. Read carefully. Decide what the problem is asking for.
2. Choose a variable. Let an unknown length or quantity be represented by a variable such as \(x\).
3. Write expressions. Use the variable to describe other lengths or quantities.
4. Use a formula. This might be area, perimeter, or the Pythagorean theorem.
5. Build an equation. This is the mathematical model.
6. Solve the quadratic equation.
7. Check the answer in the context. A negative length usually does not make sense.

2. Common Geometry Formulas Used in Quadratic Models

Here are some formulas you will use often.

• Rectangle area: \(A = lw\)
• Rectangle perimeter: \(P = 2l + 2w\)
• Triangle area: \(A = \frac{1}{2}bh\)
• Pythagorean theorem: \(a^2 + b^2 = c^2\)

If a side length is described using words like “3 more than the width” or “2 less than the length,” turn that into an algebraic expression.

For example:

• If the width is \(x\), then “length is 5 more than the width” becomes \(x + 5\).
• If one leg of a right triangle is \(x\), and the other is 4 longer, then the second leg is \(x + 4\).

3. Modeling with Area

Area problems often lead to multiplication of two expressions. Since multiplying variable expressions can produce an \(x^2\) term, these problems often become quadratic equations.

Worked Example 1: Rectangle with Known Area

A rectangle has an area of \(48\) square units. Its length is \(4\) units more than its width. Find the dimensions.

Step 1: Choose a variable.

Let the width be \(x\).

Step 2: Write the other dimension.

Then the length is \(x + 4\).

Step 3: Use the area formula.

Since area of a rectangle is \(A = lw\), we get

$$x(x+4)=48$$

Step 4: Write in quadratic form.

$$x^2 + 4x - 48 = 0$$

Step 5: Solve.

Factor the quadratic:

$$x^2 + 4x - 48 = (x+8)(x-6)=0$$

So the solutions are

$$x=-8 \quad \text{or} \quad x=6$$

Step 6: Check for reasonableness.

A width cannot be negative, so \(x=6\).

The width is \(6\) units, and the length is

$$6+4=10$$

Answer: The dimensions are \(6\) units by \(10\) units.

Important idea: Even when a quadratic has two solutions, only one may make sense in a word problem.

4. Modeling with Perimeter and Area Together

Sometimes a problem gives a perimeter and asks for dimensions with a certain area relationship. In these problems, you may first use the perimeter to write one variable in terms of the other.

Worked Example 2: Rectangle with Given Perimeter

A rectangle has perimeter \(30\) units and area \(56\) square units. Find its dimensions.

Step 1: Choose variables.

Let the length be \(l\) and the width be \(w\).

Step 2: Use the perimeter.

$$2l + 2w = 30$$

Divide by 2:

$$l + w = 15$$

Solve for one variable:

$$l = 15 - w$$

Step 3: Use the area.

$$lw = 56$$

Substitute \(l = 15 - w\):

$$w(15-w)=56$$

Step 4: Write as a quadratic.

$$15w - w^2 = 56$$ $$-w^2 + 15w - 56 = 0$$

Multiply by \(-1\) to make the leading coefficient positive:

$$w^2 - 15w + 56 = 0$$

Factor:

$$ (w-7)(w-8)=0 $$

So \(w=7\) or \(w=8\).

If \(w=7\), then \(l=8\). If \(w=8\), then \(l=7\).

Answer: The rectangle’s dimensions are \(7\) units and \(8\) units.

Notice that switching length and width does not change the rectangle.

5. Modeling with the Pythagorean Theorem

Right triangle problems often lead to quadratics because the theorem uses squares:

$$a^2+b^2=c^2$$

If one or more side lengths are written with variables, squaring them can create a quadratic equation.

Worked Example 3: Right Triangle Side Lengths

The legs of a right triangle are consecutive integers. The hypotenuse is \(13\). Find the lengths of the legs.

Step 1: Choose a variable.

Let one leg be \(x\). Since the legs are consecutive integers, the other leg is \(x+1\).

Step 2: Use the Pythagorean theorem.

$$x^2 + (x+1)^2 = 13^2$$ $$x^2 + (x+1)^2 = 169$$

Step 3: Expand.

$$x^2 + x^2 + 2x + 1 = 169$$ $$2x^2 + 2x + 1 = 169$$ $$2x^2 + 2x - 168 = 0$$

Divide by 2:

$$x^2 + x - 84 = 0$$

Step 4: Factor.

$$ (x+12)(x-7)=0 $$

So \(x=-12\) or \(x=7\).

A side length cannot be negative, so \(x=7\).

The other leg is

$$x+1=8$$

Answer: The legs are \(7\) and \(8\).

6. What Optimization Means

An optimization problem asks for the greatest or least possible value of something.

In geometry, common optimization questions include:

• What dimensions give the greatest area?
• What dimensions give the smallest perimeter?
• What is the maximum or minimum value of a measurement?

These problems often produce a quadratic expression. The key idea is that a quadratic can have a maximum or minimum value.

If the quadratic is written as

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

then:

• If \(a>0\), the parabola opens upward and has a minimum.
• If \(a<0\), the parabola opens downward and has a maximum.

The x-value of the vertex is

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

This is often how we find the best dimensions in an optimization problem.

7. Optimization with Area

Worked Example 4: Maximum Area with Fixed Perimeter

You have \(40\) meters of fencing to make a rectangular garden. What dimensions give the maximum area?

Step 1: Choose a variable.

Let the width be \(x\) meters.

Step 2: Use the perimeter.

$$2l + 2w = 40$$

Substitute \(w=x\):

$$2l + 2x = 40$$ $$2l = 40 - 2x$$ $$l = 20 - x$$

Step 3: Write the area function.

$$A = lw$$ $$A = x(20-x)$$ $$A = 20x - x^2$$

This is a quadratic:

$$A = -x^2 + 20x$$

Because the coefficient of \(x^2\) is negative, the parabola opens downward, so the area has a maximum.

Step 4: Find the vertex.

Here, \(a=-1\) and \(b=20\). So

$$x=\frac{-b}{2a}=\frac{-20}{2(-1)}=10$$

So the width is \(10\) meters.

Then the length is

$$20-10=10$$

Step 5: Find the maximum area.

$$A=10\cdot 10=100$$

Answer: The maximum area occurs when the garden is \(10\) m by \(10\) m, and the maximum area is \(100\text{ m}^2\).

Important pattern: For rectangles with a fixed perimeter, the maximum area happens when the rectangle is a square.

8. How to Decide Which Solution to Keep

After solving a quadratic in a word problem, always check whether your solution makes sense.

• Lengths must be positive.
• Areas must be nonnegative.
• A value must fit the wording of the problem.

For example, if you solve and get \(x=12\) and \(x=-3\), the negative answer is usually rejected if \(x\) represents a side length.

Also, be careful if a value makes another expression zero or negative. For instance, if length is \(15-x\), then \(x\) cannot be greater than \(15\).

9. Common Mistakes to Avoid

• Using the wrong formula. Make sure you know whether the problem involves area, perimeter, or the Pythagorean theorem.
• Forgetting to define the variable. Always say what \(x\) stands for.
• Writing expressions incorrectly. “5 more than width” means \(x+5\), not \(5x\).
• Not putting the equation in standard form. Rearranging to \(ax^2+bx+c=0\) makes solving easier.
• Keeping impossible solutions. Check whether answers make sense in the real situation.
• Stopping too early in optimization. Find both the dimension and the maximum or minimum value if the question asks for both.

10. Problem-Solving Checklist

When you see a geometry word problem that may involve a quadratic, ask yourself:

• What am I trying to find?
• What variable should I use?
• How can I write the other measurements in terms of that variable?
• Which formula connects the information?
• What quadratic equation or expression do I get?
• How do I solve it?
• Does my answer make sense in the context?

11. Quick Practice Thinking

Here are some examples of how to recognize the model you need:

• If a rectangle has one side “3 less than twice the other” and you know the area, use the area formula.
• If a right triangle has legs written with variable expressions, use the Pythagorean theorem.
• If a rectangle has a fixed perimeter and asks for greatest area, write area as a quadratic function and find the vertex.

12. Summary

Geometry word problems often become quadratic equations when you model them with algebra. Area problems usually involve multiplying expressions, and right triangle problems use squared terms from the Pythagorean theorem.

Optimization problems ask for a maximum or minimum value. In those cases, write the quantity to be optimized as a quadratic expression and use the vertex to find the best value.

The most important steps are to define a variable, write expressions carefully, choose the correct formula, solve the quadratic, and check whether the answer makes sense in the situation.

Modeling Word Problems: Kinematics and Rates

Quadratic equations are not just algebra exercises. They are also used to model real situations, especially when something changes in a curved, non-linear way.

In this lesson, you will learn how to turn word problems about motion and rates into quadratic equations. Then you will solve those equations and interpret the answers in the context of the problem.

This idea is especially important in kinematics, the study of motion, and in rate problems, where the speed of work or motion changes over time.

Big Idea: A quadratic model often appears when a quantity depends on time squared, such as height in projectile motion or distance traveled under constant acceleration.

1. Recognizing when a quadratic model is needed

Many word problems can be modeled with a linear equation, but some situations involve a changing rate. In those cases, the equation often includes a squared term like \(t^2\), which makes it quadratic.

Common signs that a quadratic model may be needed include:

• Height of an object thrown upward or dropped
• Motion with constant acceleration
• Distance formulas that include squared time
• Rate situations where a changing amount creates a second-degree equation

A quadratic equation usually has the form

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

To solve a word problem, we usually follow these steps:

1. Define the variable.
2. Write an equation from the situation.
3. Rewrite it in standard quadratic form if needed.
4. Solve by factoring, completing the square, or using the quadratic formula.
5. Check which solution makes sense in the real-world situation.

2. Quadratic models in projectile motion

One of the most common quadratic models in kinematics is the height of an object moving up and down under gravity.

In many school problems, the height in feet after \(t\) seconds is modeled by

$$h(t) = -16t^2 + v_0t + h_0$$

Here:

• \(h(t)\) is the height
• \(v_0\) is the initial upward velocity
• \(h_0\) is the initial height
• The \(-16t^2\) part shows the effect of gravity

The negative coefficient tells us the graph opens downward, which matches real life: the object rises, reaches a highest point, and then falls.

Questions about projectile motion often ask:

• How high is the object after a certain time?
• When does the object hit the ground?
• When does it reach a certain height?
• What is the maximum height?

3. Quadratic models with constant acceleration

If an object moves with constant acceleration, its distance from a starting point can often be modeled by

$$d = v_0t + \frac{1}{2}at^2$$

where:

• \(d\) is distance
• \(v_0\) is initial velocity
• \(a\) is acceleration
• \(t\) is time

This is quadratic in \(t\) because of the \(t^2\) term. If you are given a distance and asked for the time, you may need to solve a quadratic equation.

4. Quadratic equations in rate problems

Not every rate problem is quadratic, but some become quadratic after modeling.

For example, suppose the rate of work changes over time, or one quantity depends on another in a way that creates a product involving a variable. When that product expands, a squared term may appear.

A common strategy in rate problems is:

• Let a variable represent time, speed, or another unknown.
• Use a formula such as \(\text{distance} = \text{rate} \cdot \text{time}\) or a work relationship.
• Substitute the expressions from the problem.
• Simplify to get a quadratic equation.

5. Important skill: interpreting solutions

Quadratic equations can have two solutions, one solution, or no real solutions. In word problems, this matters.

For example:

• A negative time usually does not make sense.
• Two positive times may both make sense, such as when an object passes the same height once on the way up and once on the way down.
• A decimal answer may need to be rounded depending on the question.

Always go back to the original situation and ask, Does this answer make sense?

Worked Example 1: Finding when an object hits the ground

A ball is thrown upward from a height of 5 feet with an initial velocity of 48 feet per second. Its height is modeled by

$$h(t) = -16t^2 + 48t + 5$$

When does the ball hit the ground?

Step 1: Understand what “hits the ground” means.

The ball hits the ground when its height is 0. So set \(h(t)=0\):

$$-16t^2 + 48t + 5 = 0$$

Step 2: Use the quadratic formula.

For \(ax^2+bx+c=0\), the quadratic formula is

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

Here, \(a=-16\), \(b=48\), and \(c=5\).

$$t = \frac{-48 \pm \sqrt{48^2 - 4(-16)(5)}}{2(-16)}$$

$$t = \frac{-48 \pm \sqrt{2304 + 320}}{-32}$$

$$t = \frac{-48 \pm \sqrt{2624}}{-32}$$

Since \(\sqrt{2624} \approx 51.23\),

$$t = \frac{-48 \pm 51.23}{-32}$$

This gives two solutions:

$$t \approx \frac{3.23}{-32} \approx -0.10$$

$$t \approx \frac{-99.23}{-32} \approx 3.10$$

Step 3: Interpret the answers.

Negative time does not make sense here, so the meaningful answer is

$$t \approx 3.1 \text{ seconds}$$

Answer: The ball hits the ground after about \(3.1\) seconds.

Worked Example 2: Finding when an object reaches a certain height

A firework is launched from the ground. Its height after \(t\) seconds is

$$h(t) = -16t^2 + 80t$$

At what times is the firework 96 feet above the ground?

Step 1: Set the height equal to 96.

$$-16t^2 + 80t = 96$$

Step 2: Rewrite in standard form.

$$-16t^2 + 80t - 96 = 0$$

Divide every term by \(-16\):

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

Step 3: Factor.

$$ (t-2)(t-3) = 0 $$

So,

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

Step 4: Interpret the answers.

Both times make sense. The firework is at 96 feet once on the way up and once on the way down.

Answer: The firework is 96 feet high at \(2\) seconds and \(3\) seconds.

Worked Example 3: Motion with constant acceleration

A car starts with a speed of 10 meters per second and accelerates at 4 meters per second squared. How long does it take to travel 72 meters?

Use the distance formula

$$d = v_0t + \frac{1}{2}at^2$$

Step 1: Substitute the known values.

$$72 = 10t + \frac{1}{2}(4)t^2$$

$$72 = 10t + 2t^2$$

Step 2: Write in standard form.

$$2t^2 + 10t - 72 = 0$$

Divide by 2:

$$t^2 + 5t - 36 = 0$$

Step 3: Factor.

$$ (t+9)(t-4) = 0 $$

So,

$$t=-9 \quad \text{or} \quad t=4$$

Step 4: Interpret the answers.

A negative time does not fit the situation, so the answer is

$$t=4 \text{ seconds}$$

Answer: The car travels 72 meters in 4 seconds.

Worked Example 4: A quadratic rate model

A machine produces parts. After an upgrade, its production rate is \((x+4)\) parts per minute instead of \(x\) parts per minute. In 6 minutes, it makes 24 more parts than before. Find the original rate.

Step 1: Write expressions for the number of parts made.

Original number of parts in 6 minutes:

$$6x$$

New number of parts in 6 minutes:

$$6(x+4)$$

This particular setup is linear, so let us extend the situation slightly to make it quadratic in a realistic modeling way.

Suppose instead that the machine's rate after \(t\) minutes is \((x+t)\) parts per minute, starting from \(x\) parts per minute. Over \(x\) minutes, it produces a total of 60 parts. Model this with a quadratic equation.

Step 2: Build the expression.

If we use the simplified model “number of parts = rate \(\times\) time,” then total parts are

$$x(x+x) = 60$$

because time is \(x\) minutes and the changing rate expression becomes \((x+x)=2x\).

So,

$$2x^2 = 60$$

$$x^2 = 30$$

$$x = \pm \sqrt{30}$$

Step 3: Interpret the result.

A production rate cannot be negative, so

$$x = \sqrt{30} \approx 5.48$$

Answer: The original rate is about \(5.48\) parts per minute.

This example shows that even in rate situations, a quadratic equation can appear when a variable is multiplied by another expression involving that same variable.

6. Tips for setting up word problems correctly

• Choose your variable carefully. Write down what it represents.
• Look for formulas. In motion problems, common formulas include height, distance, rate, and time.
• Pay attention to units. Seconds, feet, meters, and minutes should match the formula.
• Write the equation before solving. The hardest part is often the modeling, not the algebra.
• Check for reasonable answers. Negative time or impossible distances should be rejected.

7. Common mistakes to avoid

• Forgetting to set the equation equal to 0 before using factoring or the quadratic formula
• Using the wrong value for \(a\), \(b\), or \(c\)
• Keeping a negative time answer when it does not make sense
• Missing the fact that two positive answers can both be correct in projectile problems
• Ignoring what the question is asking, such as height, time, or maximum value

8. Quick problem-solving checklist

1. What quantity am I trying to find?
2. What variable should I use?
3. What formula or relationship fits the situation?
4. What quadratic equation do I get?
5. How do I solve it?
6. Which solution(s) make sense in context?

Summary

Quadratic equations are useful for modeling real-world situations where change is not constant in a straight-line way. In kinematics, they appear in projectile motion and constant acceleration formulas because time is squared.

In rate problems, quadratics can appear when variables multiply expressions involving the same variable. The key is to translate the words into an equation, solve carefully, and then decide which answers are reasonable in the real situation.