Math Core

Some quadratic equations have no $x$ term at all, like $x^2 = 25$ or $3x^2 - 48 = 0$ . Others are a perfect square set equal to a number, like $(x - 3)^2 = 16$ . You don't need to factor these. You can undo the square directly by taking square roots, and this method also handles answers that don't factor nicely, such as $\sqrt{7}$ .

Two square roots, not one

Which numbers have a square of $25$ ? Both $5$ and $-5$ , since $5^2 = 25$ and $(-5)^2 = 25$ . So the equation $x^2 = 25$ has two solutions. The symbol $\sqrt{25}$ means only the positive root, $5$ , so to capture both you write

x = \pm\sqrt{25} = \pm 5,

read "plus or minus five."

Solving x² = d

If $d > 0$ , then $x^2 = d$ has two solutions: $x = \sqrt{d}$ and $x = -\sqrt{d}$ , written $x = \pm\sqrt{d}$ .
If $d = 0$ , the only solution is $x = 0$ .
If $d < 0$ , there is no real solution, because no real number squared is negative.

Isolate the square first

Before taking a square root, get the squared expression alone on one side, exactly as you would isolate $x$ in a linear equation: undo addition and subtraction, then multiplication and division.

Worked example: Isolating x²

Solve each equation.

$3x^2 - 48 = 0$
$2x^2 + 5 = 23$

Solutions.

Add $48$ : $3x^2 = 48$ . Divide by $3$ : $x^2 = 16$ . So $x = \pm 4$ .
Subtract $5$ : $2x^2 = 18$ . Divide by $2$ : $x^2 = 9$ . So $x = \pm 3$ .

Common mistake

Don't take the square root before the square is alone. For $x^2 + 9 = 25$ , writing $x + 3 = 5$ is wrong, because $\sqrt{x^2 + 9}$ is not $x + 3$ . Subtract first: $x^2 = 16$ , so $x = \pm 4$ . And don't forget the $\pm$ : dropping it throws away half of the answers.

Irrational solutions

When $d$ is not a perfect square, leave the answer as a simplified radical. You simplified radicals in the exponents unit by pulling out perfect-square factors: $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$ .

Worked example: A radical answer

Solve $4x^2 - 7 = 13$ . Give exact answers, then approximate them.

\begin{aligned} 4x^2 &= 20 \\ x^2 &= 5 \\ x &= \pm\sqrt{5} \end{aligned}

The exact solutions are $\sqrt{5}$ and $-\sqrt{5}$ , which are approximately $2.24$ and $-2.24$ .

Squared binomials

The same idea works when a whole binomial is squared. Treat the binomial as a single block, take the square root of both sides, and then solve the two linear equations that result.

Worked example: Taking the root of a binomial square

Solve $(x - 3)^2 = 16$ and $(x + 2)^2 = 7$ .

For the first, $x - 3 = \pm 4$ :

x - 3 = 4 \ \Longrightarrow\ x = 7 \qquad\qquad x - 3 = -4 \ \Longrightarrow\ x = -1.

For the second, $x + 2 = \pm\sqrt{7}$ , so $x = -2 \pm \sqrt{7}$ . That's two solutions: $-2 + \sqrt{7} \approx 0.65$ and $-2 - \sqrt{7} \approx -4.65$ .

This is exactly how you find the x-intercepts of a parabola in vertex form. The equation $2(x - 1)^2 - 18 = 0$ asks where $y = 2(x - 1)^2 - 18$ crosses the $x$ -axis.

Worked example: Zeros from vertex form

Solve $2(x - 1)^2 - 18 = 0$ .

\begin{aligned} 2(x - 1)^2 &= 18 \\ (x - 1)^2 &= 9 \\ x - 1 &= \pm 3 \\ x = 4 \quad &\text{or} \quad x = -2 \end{aligned}

The zeros are $3$ units on either side of the axis of symmetry $x = 1$ .

y = 2(x - 1)² - 18 crosses the x-axis at -2 and 4, each 3 units from the axis x = 1.Open in grapher →

When there is no solution

Try to solve $x^2 + 4 = 0$ . Subtracting gives $x^2 = -4$ , and no real number has a negative square. The equation has no real solution. On the graph, the parabola $y = x^2 + 4$ has its vertex at $(0, 4)$ and opens up, so it never reaches the $x$ -axis.

y = x² + 4 never touches the x-axis, so x² + 4 = 0 has no real solutions.Open in grapher →

Tip

Only keep the answers that make sense in context. A square with area $50$ square feet has side length $\sqrt{50} = 5\sqrt{2} \approx 7.07$ feet. The equation $s^2 = 50$ also has the solution $-5\sqrt{2}$ , but a length can't be negative.

Practice

Practice 1

Solve $x^2 = 81$ .

Separate answers with commas, e.g. 2, -5

Practice 2

Solve $5x^2 - 80 = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 3

The equation $x^2 + 6 = 30$ has two solutions. What is the positive solution? Give an exact answer, such as 3sqrt(2).

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Solve $(x - 5)^2 = 36$ .

Separate answers with commas, e.g. 2, -5

Practice 5

Solve $(x + 4)^2 = 3$ .

Practice 6

How many real solutions does $x^2 + 10 = 1$ have?

Practice 7

Solve $3(x + 1)^2 - 12 = 63$ .

Separate answers with commas, e.g. 2, -5

Practice 8

A stone is dropped from a cliff $100$ feet high. Its height after $t$ seconds is $h(t) = -16t^2 + 100$ . After how many seconds does it reach the ground?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.