⚔️ Algebra Kingdom · Algebra

Quadratic Equations

Solve equations containing a squared term by factoring, by taking square roots, and with the quadratic formula.

In short

A quadratic equation normally has two solutions, because a parabola can cross the horizontal axis twice.
Taking a square root produces both a positive and a negative answer: $x 2 = 49$ means $x = 7$ or $x = -7$ .
The zero-product rule only works when one side is 0, so rearrange before you factor.
The quadratic formula $x =$ (-b $+-$ sqrt(b^2 - $4ac)) / (2a)$ always works, and the discriminant $b 2$ - 4ac says how many real solutions there are.
In a word problem, solve fully first and then reject any root the situation forbids, giving the reason.

Why there are usually two answers

A quadratic equation contains a squared term and nothing higher: $ax 2 +$ bx $+ c = 0$ , with a not equal to 0.

Linear equations have one solution because the graph of $y = 2x + 7$ is a straight line and a line crosses the horizontal axis once. The graph of $y = x 2$ - $5x + 6$ is a parabola, a U-shaped curve, and $a U$ can cross the axis twice. Those two crossing points are the two solutions.

Squaring is what causes it. Both 4 and -4 square to 16, so both are candidates whenever a square is undone.

A parabola can also touch the axis once (one repeated solution) or miss it entirely (no real solutions). But the default expectation is two, and losing one is the single most common error in this topic.

Taking square roots

When the equation can be arranged so that a square sits alone on one side, take the square root of both sides — and keep both signs.

$x 2 = 49$ gives $x = 7$ or $x = -7$ , usually written $x = +-7$ .

The same works when a whole bracket is squared:

(x - $3) 2 = 16$
x - $3 = 4$ or x - $3 = -4$
$x = 7$ or $x = -1$

If there is a multiplier, divide it away first: 5(x - $2) 2 = 45$ becomes (x - $2) 2 = 9$ , then x - $2 = +-3$ , so $x = 5$ or $x = -1$ .

One warning: $x 2 = 49$ does not mean $x = 49 / 2$ . Undoing a square is a square root, not a halving.

The zero-product rule

This is the idea that makes factoring so powerful:

If a product of two numbers is 0, then at least one of them must be 0.

Nothing else has this property. If a product equals 12 you learn very little, because 12 can be split thousands of ways. If it equals 0, one factor is pinned down completely.

So to solve a quadratic by factoring:

Rearrange until one side is 0. This step is not optional.
Factor the other side.
Set each bracket equal to 0 and solve those small equations.

For $x 2$ - $5x + 6 = 0$ : factor to (x - 2)(x - $3) = 0$ , so x - $2 = 0$ or x - $3 = 0$ , giving $x = 2$ or $x = 3$ .

Watch the signs when reading off the answers: the bracket $(x + 4)$ gives $x =$ -4, not 4.

Also never divide both sides by x to tidy an equation like $x 2$ - $7x = 0$ . Dividing by x throws away the perfectly good solution $x = 0$ . Factor instead: x(x - $7) = 0$ , so $x = 0$ or $x = 7$ .

The quadratic formula

Some quadratics do not factor over the whole numbers. The formula always works:

$x =$ (-b $+-$ sqrt(b^2 - $4ac)) / (2a)$

Read $a, b$ and c off the equation once it is written as $ax 2 +$ bx $+ c = 0$ , taking the signs with them. In $2x 2$ - $7x + 3 = 0, a = 2, b = -7$ and $c = 3$ .

The part under the square root, $b 2$ - 4ac, is called the discriminant, and it forecasts the answer before you finish:

positive: two different real solutions (the parabola crosses the axis twice)
zero: one repeated solution (the parabola just touches the axis)
negative: no real solutions (the parabola misses the axis)

Two things to be careful about. The whole numerator is divided by 2a, not just the square root. And -b means the opposite of b, so if $b = -7$ then -b $= +7$ .

Quadratics that describe something real

Areas and falling objects are naturally quadratic, because both involve a quantity multiplied by itself or by time squared.

A yard that is 3 metres longer than it is wide with an area of 40 square metres gives w(w $+ 3) = 40$ , so $w 2 + 3w$ - $40 = 0$ and (w - $5)(w + 8) = 0$ . The algebra offers $w = 5$ and $w = -8$ .

Here comes the step that only exists in word problems: check both answers against the situation. A width of -8 metres is meaningless, so the yard is 5 metres wide. The algebra is not wrong to produce it — the equation simply knows nothing about yards.

The same happens with height: if $h = -5t 2 + 15t + 20$ gives $t = 4$ or $t = -1$ , then $t = -1$ is a time before the flare was fired and is rejected.

So the routine is: solve fully, find both roots, then discard the ones the story forbids. Never discard a root just because it is inconvenient — say why it cannot happen.

Worked examples

Example 1

Solve $x 2$ - 5x - $24 = 0$ .

One side is already 0, so factoring will work.
Look for two numbers with product -24 and sum -5. The product is negative, so the numbers have opposite signs.
Factor pairs of $24: 1$ and $24, 2$ and $12, 3$ and $8, 4$ and 6. The pair 3 and 8 differ by 5, so try -8 and $+3$ .
$(-8) x 3 = -24$ and $-8 + 3 = -5$ , so $x 2$ - 5x - $24 = (x$ - $8)(x + 3)$ .
Zero-product rule: x - $8 = 0$ or $x + 3 = 0$ , so $x = 8$ or $x = -3$ .
Check $x = -3: 9 + 15$ - $24 = 0$ . Check $x = 8: 64$ - 40 - $24 = 0$ . Both work.

Example 2

Solve $2x 2$ - $7x + 3 = 0$ using the quadratic formula.

Read off the coefficients with their signs: $a = 2, b = -7, c = 3$ .
Discriminant: $b 2$ - $4ac = (-7) 2$ - $4 x 2 x 3 = 49$ - $24 = 25$ .
25 is positive, so there are two real solutions, and sqrt(25) $= 5$ .
$x = (7 +- 5) / (2 x 2) = (7 +- 5) / 4$ .
$x = 12 / 4 = 3$ , or $x = 2 / 4 = 0.5$ .
Check $x = 0.5: 2 x 0.25$ - $3.5 + 3 = 0.5$ - $3.5 + 3 = 0$ . Correct.

Example 3

A rectangular banner is 4 cm taller than it is wide and covers 96 square centimetres. How wide is it?

Let w be the width in centimetres. Then the height is $w + 4$ .
Area: w(w $+ 4) = 96$ , so $w 2 + 4w$ - $96 = 0$ .
Two numbers with product -96 and sum $4: 12$ and -8. So $(w + 12)(w$ - $8) = 0$ .
$w = -12$ or $w = 8$ .
A width cannot be negative, so $w = 8$ . The banner is 8 cm wide and 12 cm tall.
Check: $8 x 12 = 96$ square centimetres.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Solve for $x. x 2 = 49$ Enter the smaller root first, as $(a, b)$ . Fractions or decimals are both fine.

Answer: $(-7, 7)$

$x 2 = 49$
Take the square root of both sides: $x = 7$ or $x = -7$ .
Smaller root first: $(-7, 7)$ .

Problem 2

Difficulty 3 of 5

Solve for $x. x 2$ - 7x - $8 = 0$ Enter the smaller root first, as $(a, b)$ . Fractions or decimals are both fine.

Answer: $(-1, 8)$

$x 2$ - 7x - $8 = 0$
Two numbers with product -8 and sum $-7: 1$ and -8.
Factor: $(x + 1)(x$ - $8) = 0$
By the zero-product rule, $(x + 1) = 0$ or (x - $8) = 0$ .
$x = -1$ or $x = 8$ , so the answer is $(-1, 8)$ .
Check $x = -1: (-1) 2$ - $7 x (-1)$ - $8 = 0$ .

Problem 3

Difficulty 4 of 5

Solve for $x. 2x 2$ - $21x + 54 = 0$ Enter the smaller root first, as $(a, b)$ . Fractions or decimals are both fine.

Answer: $(4.5, 6)$

$2x 2$ - $21x + 54 = 0$
Factor: (2x - 9)(x - $6) = 0$
Zero-product rule: (2x - $9) = 0$ gives $2x = 9$ , so $x = 4.5$ .
(x - $6) = 0$ gives $x = 6$ .
Smaller root first: $(4.5, 6)$ .

Common mistakes

Giving only one root — usually only the positive square root, or only one of the two brackets.
Reading the roots straight out of the brackets: saying $(x + 3)(x$ - $5) = 0$ gives $x = 3$ and $x = -5$ instead of $x = -3$ and $x = 5$ .
Factoring while the equation still equals something other than 0, which makes the zero-product rule useless.
Dividing both sides by x in an equation like $x 2 = 7x$ , which silently loses the solution $x = 0$ .
In the quadratic formula, dividing only the square root by 2a instead of the whole numerator, or forgetting that -b is positive when b is negative.

What you should be able to do

Solve a quadratic by factoring and the zero-product property.
Solve $x 2 = k$ by taking square roots, keeping both signs.
Apply the quadratic formula and interpret the discriminant.
Solve projectile and area word problems with quadratics.

Where this fits in the curriculum

Common Core

HSA-REI.B.4.B
High school — Solve quadratic equations by inspection, taking square roots, completing the square, the quadratic formula and factoring.
HSA-SSE.B.3.A
High school — Factor a quadratic to reveal its zeros, using the zero-product property.
HSA-CED.A.1
High school — Create equations in one variable, including quadratics, and use them to solve problems.

SAT

Passport to Advanced Math
Solving quadratic equations and interpreting their solutions.