How to solve quadratic equations: 3 methods

Q: Which method should I use to solve a quadratic?

If there is no x term, or you see a squared group like (x - 2)^2, use square roots. If the numbers factor easily, factor. Otherwise use the quadratic formula, which always works. Completing the square also always works and is how the formula is built.

Q: What does the discriminant tell you?

The discriminant is b^2 - 4ac, the part under the square root in the formula. Positive means two real solutions, zero means exactly one, and negative means none, because you cannot take the square root of a negative number with real numbers.

The key idea

Factoring works because of the zero product property: if two numbers multiply to 0, at least one of them is 0. That is why the equation must equal zero before you factor.

(x + 7)(x - 4) = 0 \;\Longrightarrow\; x + 7 = 0 \;\text{ or }\; x - 4 = 0

When factoring is hard or impossible, the quadratic formula solves any $ax^2 + bx + c = 0$ :

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

Which method?
The equation looks like	Use	Example
no x term, or a squared group	square roots	$3(x - 2)^2 = 75$
easy whole-number factors	factoring	$x^2 + 3x - 28 = 0$
anything else	quadratic formula	$2x^2 - 4x - 3 = 0$

The discriminant b^2 - 4ac
If it is	Real solutions	The graph $y = ax^2 + bx + c$
positive	two	crosses the x-axis twice
zero	one (a double root)	touches the x-axis once
negative	none	never reaches the x-axis

Worked examples

Example 1: factoring

Problem Solve $x^2 + 3x = 28$ .

Get zero on one side first. Subtract 28.
$x^2 + 3x - 28 = 0$
Find two numbers that multiply to $-28$ and add to 3: they are 7 and $-4$ .
$(x + 7)(x - 4) = 0$
Set each factor equal to zero.
$x + 7 = 0 \quad\text{or}\quad x - 4 = 0$
Solve each one.
$x = -7 \quad\text{or}\quad x = 4$

Answer $x = -7$ or $x = 4$

Example 2: square roots

Problem Solve $3(x - 2)^2 = 75$ .

Divide both sides by 3 to get the squared group alone.
$(x - 2)^2 = 25$
Take the square root of both sides. Both 5 and $-5$ square to 25, so write both.
$x - 2 = 5 \quad\text{or}\quad x - 2 = -5$
Add 2 in each case.
$x = 7 \quad\text{or}\quad x = -3$

Answer $x = 7$ or $x = -3$

Example 3: the quadratic formula

Problem Solve $2x^2 - 4x - 3 = 0$ .

Here $a = 2$ , $b = -4$ , $c = -3$ . Work out the discriminant first. Note that $(-4)^2$ is positive 16.
$(-4)^2 - 4(2)(-3) = 16 + 24 = 40$
Put everything into the formula. $-b$ is $-(-4) = 4$ , and $2a = 4$ .
$x = \frac{4 + \sqrt{40}}{4} \quad\text{or}\quad x = \frac{4 - \sqrt{40}}{4}$
Simplify: $\sqrt{40} = 2\sqrt{10}$ , then divide every term on top by the 4 (that is, divide top and bottom by 2).
$x = \frac{2 + \sqrt{10}}{2} \quad\text{or}\quad x = \frac{2 - \sqrt{10}}{2}$

Answer $x = \frac{2 \pm \sqrt{10}}{2}$ , about 2.58 and $-0.58$

Example 4 (test-hard): a word problem with one answer that makes sense

Problem A ball is thrown upward from a 4-foot platform. Its height in feet after t seconds is $h = -16t^2 + 48t + 4$ . When does it hit the ground? Round to the nearest hundredth of a second.

The ground is height 0.
$-16t^2 + 48t + 4 = 0$
Divide every term by $-4$ to make the numbers smaller and a positive.
$4t^2 - 12t - 1 = 0$
It does not factor, so use the formula with $a = 4$ , $b = -12$ , $c = -1$ . Discriminant:
$(-12)^2 - 4(4)(-1) = 144 + 16 = 160$
Formula: $t = \frac{12 \pm \sqrt{160}}{8}$ . Since $\sqrt{160} = 4\sqrt{10}$ , divide top and bottom by 4.
$t = \frac{3 + \sqrt{10}}{2} \quad\text{or}\quad t = \frac{3 - \sqrt{10}}{2}$
The second value is about $-0.08$ . Negative time is before the throw, so it does not fit the story. The first is about 3.08.

Answer About 3.08 seconds, which is $t = \frac{3 + \sqrt{10}}{2}$ .

Common mistakes

Factoring before setting the equation to zero. $x(x + 3) = 28$ does not mean $x = 28$ . The zero product property only works with 0 on one side.
Forgetting the negative root. $(x - 2)^2 = 25$ gives $x - 2 = 5$ or $x - 2 = -5$ .
Dividing both sides by x. From $x^2 = 5x$ , dividing by x loses the answer $x = 0$ . Fix: move everything to one side and factor: $x(x - 5) = 0$ .
Sign errors with a negative b. If $b = -4$ , then $-b = 4$ and $b^2 = 16$ . Fix: write $-(-4)$ and $(-4)^2$ with parentheses.
Dividing only part of the top by 2a. The fraction bar covers the whole $-b \pm \sqrt{b^2 - 4ac}$ .

Quick methods

Practice

5 practice questions

Solve $x^2 - 5x - 14 = 0$ .
1. $x = 7$ or $x = -2$
2. $x = -7$ or $x = 2$
3. $x = 14$ or $x = -1$
4. $x = 5$ or $x = -14$
Show answer
Answer: $x = 7$ or $x = -2$
Factor: $(x - 7)(x + 2) = 0$ , since $-7 \cdot 2 = -14$ and $-7 + 2 = -5$ . Then $x = 7$ or $x = -2$ . The choice $-7$ or 2 takes the numbers from the factors without changing their signs.
Solve $x^2 = 6x$ .
1. $x = 6$
2. $x = 0$ or $x = 6$
3. $x = 0$ or $x = -6$
4. $x = \sqrt{6}$
Show answer
Answer: $x = 0$ or $x = 6$
Move everything to one side: $x^2 - 6x = 0$ , so $x(x - 6) = 0$ . Both 0 and 6 work. Dividing both sides by x gives only 6 and loses the 0.
Solve $(x + 1)^2 = 49$ .
1. $x = 6$
2. $x = 6$ or $x = -8$
3. $x = 48$
4. $x = 8$ or $x = -6$
Show answer
Answer: $x = 6$ or $x = -8$
Square root both sides: $x + 1 = 7$ or $x + 1 = -7$ . So $x = 6$ or $x = -8$ . Just 6 misses the negative root, and 8 or $-6$ adds 1 instead of subtracting it.
How many real solutions does $3x^2 - 2x + 5 = 0$ have?
1. 0
2. 1
3. 2
4. infinitely many
Show answer
Answer: 0
The discriminant is $(-2)^2 - 4(3)(5) = 4 - 60 = -56$ . It is negative, so there are no real solutions: the parabola never reaches the x-axis.
Solve $x^2 + 4x - 1 = 0$ .
1. $x = -2 \pm \sqrt{5}$
2. $x = 2 \pm \sqrt{5}$
3. $x = -2 \pm 2\sqrt{5}$
4. $x = -4 \pm \sqrt{5}$
Show answer
Answer: $x = -2 \pm \sqrt{5}$
Formula: $x = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$ . The choice $-2 \pm 2\sqrt{5}$ divides the $-4$ by 2 but not the root. $2 \pm \sqrt{5}$ forgets the minus in $-b$ .

Frequently asked questions

Which method should I use to solve a quadratic?

If there is no x term, or you see a squared group like $(x - 2)^2$ , use square roots. If the numbers factor easily, factor. Otherwise use the quadratic formula, which always works. Completing the square also always works and is how the formula is built.

What does the discriminant tell you?

The discriminant is $b^2 - 4ac$ , the part under the square root in the formula. Positive means two real solutions, zero means exactly one, and negative means none, because you cannot take the square root of a negative number with real numbers.

Why do quadratic equations have two answers?

The graph of a quadratic is a U-shaped parabola, and a U can cross a horizontal line in two places. Algebraically, both a number and its opposite square to the same value. Sometimes the two answers are the same (one solution), and sometimes there are none.

What if I get a negative number under the square root?

Then the equation has no real solutions. In Algebra 1 you write "no real solution." In Algebra 2 you learn imaginary numbers, which give two complex solutions instead. Double-check that $b^2$ was positive and the signs of a and c are right first.