A. Quadratic Functions · Quadratic Expressions and Equations

Solving Quadratic Equations by Factoring

A1.4About 70 minutes

Learning goals

I can rearrange a quadratic equation so that one side is 0.
I can solve quadratic equations by factoring and the zero product property.
I can verify roots and reject roots that do not make sense in context.

Words to know

quadratic equation
An equation that can be written ax² + bx + c = 0 with a ≠ 0.

root
A solution of an equation.

zero product property
If A × B = 0, then A = 0 or B = 0 (or both).

double root
A root that appears twice, as in (x − 4)² = 0, whose only root is 4.

verify
Substitute a root into the original equation to check that both sides are equal.

From expressions to equations

For four lessons you have been rewriting expressions. Now we use that skill to solve equations. A quadratic equation can be written ax² + bx + c = 0 with a ≠ 0. Its solutions are called roots.

Factoring solves many quadratic equations in a few lines, thanks to one simple fact about zero.

If a product is zero, at least one factor must be zero. No other number has this property: if A × B = 6, A could be 1, 2, 0.5, −3, ...

Solving by factoring: five steps

Rearrange so that one side is 0 (expand first if needed).
Factor the other side fully.
Set each factor equal to 0.
Solve each small equation.
Verify by substituting into the original equation, and reject roots that make no sense in context.

Step 1 is not optional. The zero product property works only with zero.

Worked example

Example 1: a ≠ 1

Solve 2x² − 7x − 15 = 0.

One side is already 0. Decomposition: ac = −30, sum −7: −10 and 3.
2x² − 10x + 3x − 15 = 2x(x − 5) + 3(x − 5) = (x − 5)(2x + 3).
(x − 5)(2x + 3) = 0, so x − 5 = 0 or 2x + 3 = 0.
x = 5 or x = −3/2.
Verify x = 5: 2(25) − 35 − 15 = 0 ✓. Verify x = −1.5: 2(2.25) + 10.5 − 15 = 4.5 + 10.5 − 15 = 0 ✓.

x = 5 or x = −1.5

Worked example

Example 2: The equation is not equal to 0

Solve (x + 2)(x − 3) = 14.

Tempting but wrong: "x + 2 = 14 or x − 3 = 14". The product is 14, not 0, so the zero product property does not apply.
Expand: x² − x − 6 = 14. Rearrange: x² − x − 20 = 0.
Factor: (x − 5)(x + 4) = 0, so x = 5 or x = −4.
Verify in the original: x = 5 gives (7)(2) = 14 ✓; x = −4 gives (−2)(−7) = 14 ✓.

x = 5 or x = −4

The roots of x² − x − 20 = 0 (Example 2) are the x-intercepts of y = x² − x − 20. The lesson Factors and x-Intercepts studies this connection.

Worked example

Example 3: The curriculum's equation (from A1.8)

Solve 100x² = 115x + 35.

Rearrange: 100x² − 115x − 35 = 0.
Common factor 5: 5(20x² − 23x − 7) = 0. Dividing both sides by 5: 20x² − 23x − 7 = 0.
Decomposition: ac = −140, sum −23: −28 and 5. 20x² − 28x + 5x − 7 = 4x(5x − 7) + 1(5x − 7) = (5x − 7)(4x + 1).
5x − 7 = 0 gives x = 7/5 = 1.4; 4x + 1 = 0 gives x = −1/4 = −0.25.
Verify x = 1.4: LS = 100(1.96) = 196; RS = 115(1.4) + 35 = 161 + 35 = 196 ✓. Verify x = −0.25: LS = 6.25; RS = −28.75 + 35 = 6.25 ✓.

x = 1.4 or x = −0.25

Worked example

Example 4: A problem in context

In Representing Situations with Quadratic Expressions you found that the area of a frame of width x around a 30 cm by 20 cm photo is 4x² + 100x. What frame width makes the frame's area equal to the photo's area, 600 cm²?

Equation: 4x² + 100x = 600.
Rearrange and divide by 4: 4x² + 100x − 600 = 0, so x² + 25x − 150 = 0.
Factor: product −150, sum 25: 30 and −5. (x + 30)(x − 5) = 0.
x = −30 or x = 5. A width cannot be negative, so reject x = −30.
Check: outer frame 40 cm by 30 cm = 1200 cm²; minus the photo, 600 cm² ✓.

The frame should be 5 cm wide.

Three traps

Never divide both sides by x. In x² = 6x, dividing by x gives x = 6 and loses the root x = 0. Instead: x² − 6x = 0, x(x − 6) = 0, x = 0 or 6.
A perfect square gives one root. 9x² + 12x + 4 = 0 is (3x + 2)² = 0, so x = −2/3 only (a double root).
Dividing by a number is fine (as in Example 3), because a number is never 0.

Just for fun

The zero product property is the only time in life when one zero can make the whole team's score zero, and everybody's happy about it.

Hands-on activity

Roots on the floor

You need: masking tape (or string), a measuring tape and a few coins or small objects to use as markers.

Make a number line on the floor with tape: mark −6 to 6 using 30 cm per unit.
Solve x² + 3x − 10 = 0 by factoring. Place a coin on each root.
Stand on each coin in turn while a helper calculates x² + 3x − 10 for your position. They should get 0 each time.
Stand at x = 0, x = 1 and x = −3. What values do you get? Are they positive or negative?

What to look for: (x + 5)(x − 2) = 0 gives roots x = −5 and x = 2. At those points the value is 0. At x = 0 the value is −10, at x = 1 it is −6, and at x = −3 it is −10: between the roots the expression is negative, and outside them it is positive. That is exactly what the parabola y = x² + 3x − 10 does: it dips below the x-axis between its zeros.

Your turn

Solve x² − 5x − 24 = 0.

Solve 3x² = 12x.

Solve 4x² − 9 = 0.

Solve 2x² + x − 6 = 0.

Solve x² + 16 = 8x.

Solve x(x + 4) = 21.

Solve 6x² − 7x − 3 = 0.

More practice: warm-up

Solve (x − 4)(x + 9) = 0.

Solve x(2x − 5) = 0.

Solve x² − 49 = 0.

Solve x² + 7x + 10 = 0.

Solve (3x − 1)² = 0.

More practice: on your own

Solve x²/2 − x = 4.

Solve (2x − 1)(x + 2) = 7.

Solve 5x² − 20x − 60 = 0.

A rectangle's length is 3 m more than its width, and its area is 40 m². Find its dimensions.

Sam solved x² = 6x by dividing both sides by x and got x = 6. What did Sam miss?

Challenge

Solve 12x² + 5x = 2.

The product of two consecutive positive odd integers is 143. Find them.

Write a quadratic equation with integer coefficients whose roots are 3 and −2/5.

Check your understanding

Quiz

Solve x² + 2x − 15 = 0.

Why must one side be 0 before you use the factors?

What are the roots of 3x(x − 7) = 0?

Solve x² = 10x.

Solve 2x² − 3x − 2 = 0.

A frame problem gives x = 4 or x = −29 for the width in cm. The answer is:

For parents and teachers

Big idea (A1.4): solving quadratic equations by choosing and applying a factoring strategy, including equations that must be rearranged first, like 100x² = 115x + 35 (the sample equation the curriculum gives under A1.8).

How to help: insist on the verification step: substitute each root into the original equation. It catches nearly every error, and it is a habit worth marks on tests.

Common difficulties:

Setting factors equal to a non-zero number, as in (x + 2)(x − 3) = 14 → x + 2 = 14.
Dividing by x and losing the root 0.
Sign errors when solving each factor (x + 5 = 0 gives x = −5, not 5).
Forgetting to reject roots that make no sense in context (negative widths or times).