Solving Quadratics with Square Roots

Solving quadratics with square roots works when the equation has a squared term and no x-term, like $x^2 = 25$ or $(x - 3)^2 = 16$ . Isolate the squared part, then take the square root of both sides. Use $\pm$ because both a positive and a negative number square to the same value.

Before this lesson: You should know square roots, simplifying square roots and solving quadratic equations by factoring.

Why Square Roots Undo Squares

Squaring and taking a square root are inverse operations. Each undoes the other. To solve $x^2 = 25$ , ask which numbers square to 25.

There are two: $5^2 = 25$ and $(-5)^2 = 25$ . So $x = \pm 5$ . The symbol $\pm$ means plus or minus. Forgetting the negative answer is the most common mistake.

Four rows of steps. Start with 2(x − 3)² = 32. Divide by 2 to isolate the square: (x − 3)² = 16. Take the square root of both sides, plus or minus: x − 3 = ±4. Add 3: x = 7 or x = −1. — Isolate the square first. Then the ± sign gives two equations: x − 3 = 4 and x − 3 = −4.

The Square Root Property

\text{If } x^2 = k \text{ and } k \ge 0, \text{ then } x = \pm\sqrt{k}.

Value of k	Solutions
$k \gt 0$	Two: $\sqrt{k}$ and $-\sqrt{k}$
$k = 0$	One: $x = 0$
$k \lt 0$	No real solutions, since no real number squares to a negative

Steps for solving with square roots

Isolate the squared part: $x^2$ or $(x - h)^2$ .
Take the square root of both sides. Write $\pm$ .
Simplify the radical.
If there was a number inside the parentheses, solve the two small equations.
Check both answers.

Worked Examples

Example 1: A perfect square

Solve $x^2 = 49$ .

Step 1. Take the square root of both sides. A positive number has two square roots, one positive and one negative, so write $\pm$ .

x = \pm\sqrt{49}

Step 2. $7^2 = 49$ and $(-7)^2 = 49$ .

x = \pm 7

Answer: $x = 7$ or $x = -7$

Example 2: Isolate first

Solve $3x^2 - 75 = 0$ .

Step 1. Get $x^2$ by itself. Add 75 to both sides (see two-step equations).

3x^2 = 75

Step 2. Divide both sides by 3.

x^2 = 25

Step 3. Take the square root of both sides.

x = \pm\sqrt{25}

Step 4. Simplify.

x = \pm 5

Answer: $x = 5$ or $x = -5$

Example 3: Simplify the radical

Solve $x^2 - 18 = 0$ .

Step 1. Add 18 to both sides.

x^2 = 18

Step 2. Take the square root of both sides.

x = \pm\sqrt{18}

Step 3. Simplify the radical (see simplifying square roots). The largest perfect square that divides 18 is 9.

x = \pm\sqrt{9 \cdot 2}

Step 4. Take the square root of 9, which is 3.

x = \pm 3\sqrt{2}

Step 5. To estimate, use $\sqrt{2} \approx 1.414$ .

3 \cdot 1.414 = 4.242

Answer: $x = \pm 3\sqrt{2}$ , or about $\pm 4.24$ . Leave the exact answer in simplest radical form.

Example 4: A squared binomial

Solve $(x - 4)^2 = 9$ .

Step 1. The square is already alone. Take the square root of both sides.

x - 4 = \pm\sqrt{9}

Step 2. Simplify the root.

x - 4 = \pm 3

Step 3. Add 4 to both sides.

x = 4 \pm 3

Step 4. Split into the $+$ and $-$ cases.

x = 4 + 3 \text{ or } x = 4 - 3

Step 5. Add and subtract.

x = 7 \text{ or } x = 1

Answer: $x = 7$ or $x = 1$ . When the number is not a perfect square, keep the radical. For $(x + 2)^2 = 5$ , $x + 2 = \pm\sqrt{5}$ , so $x = -2 \pm \sqrt{5}$ .

Example 5: Word problem

A rock falls from a 144-foot cliff. Its height after $t$ seconds is $h = -16t^2 + 144$ . When does it hit the ground?

Step 1. It hits the ground when $h = 0$ .

0 = -16t^2 + 144

Step 2. Add $16t^2$ to both sides.

16t^2 = 144

Step 3. Divide both sides by 16.

t^2 = 9

Step 4. Take the square root of both sides.

t = \pm\sqrt{9}

Step 5. Simplify.

t = \pm 3

Step 6. Time can’t be negative, so $t = 3$ .

Answer: The rock hits the ground after 3 seconds.

Check: $-16(3)^2 + 144 = -144 + 144 = 0$ .

Common Mistakes

Mistake	Example of the error	Fix
Forgetting the negative root	$x^2 = 36$ , so $x = 6$	$x = \pm 6$ . Both 6 and −6 square to 36.
Taking the root before isolating	$x^2 + 9 = 25$ , so $x + 3 = 5$	Subtract 9 first: $x^2 = 16$ , so $x = \pm 4$ .
Not taking the root of the right side	$(x - 4)^2 = 9$ , so $x - 4 = \pm 9$	$\sqrt{9} = 3$ , so $x - 4 = \pm 3$ .
Square root of a negative	$x^2 = -16$ , so $x = \pm 4$	$(\pm 4)^2 = 16$ , not −16. There are no real solutions.
Not simplifying	$x = \pm\sqrt{18}$	Simplify: $\pm 3\sqrt{2}$ .

When to Use the Square Root Method

Equation	Best method
$x^2 = 20$	Square roots: no x-term
$(2x + 1)^2 = 49$	Square roots: the square is already isolated
$x^2 + 6x + 8 = 0$	Factoring: it has an x-term
$x^2 + 6x = 4$	Completing the square (next lesson) or the quadratic formula

Completing the square turns any quadratic into the form $(x - h)^2 = k$ , so this method works for every quadratic. That is the next lesson.

Practice Problems

Solve $x^2 = 81$ .
Solve $x^2 - 100 = 0$ .
Solve $4x^2 = 36$ .
Solve $x^2 = 12$ .
Solve $5x^2 - 40 = 0$ .
Solve $(x + 1)^2 = 25$ .
Solve $(x - 6)^2 = 7$ .
Solve $x^2 + 10 = 1$ .
Solve $2(x - 3)^2 = 50$ .
A square garden has an area of 75 square feet. Find the length of a side in simplest radical form.

Show answers

Take the square root of both sides: $x = \pm\sqrt{81}$
Simplify: $x = \pm 9$
Add 100 to both sides: $x^2 = 100$
Take the square root of both sides: $x = \pm 10$
Divide both sides by 4: $x^2 = 9$
Take the square root of both sides: $x = \pm 3$
Take the square root of both sides: $x = \pm\sqrt{12}$
The largest perfect square that divides 12 is 4: $x = \pm\sqrt{4 \cdot 3}$
Simplify: $x = \pm 2\sqrt{3}$
Add 40 to both sides: $5x^2 = 40$
Divide both sides by 5: $x^2 = 8$
Take the square root of both sides: $x = \pm\sqrt{8}$
Simplify: $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$ : $x = \pm 2\sqrt{2}$
Take the square root of both sides: $x + 1 = \pm 5$
Subtract 1 from both sides: $x = -1 \pm 5$
Split into two cases: $x = -1 + 5 \text{ or } x = -1 - 5$
Simplify: $x = 4 \text{ or } x = -6$
Take the square root of both sides. 7 is not a perfect square, so keep the radical: $x - 6 = \pm\sqrt{7}$
Add 6 to both sides: $x = 6 \pm \sqrt{7}$
Subtract 10 from both sides: $x^2 = -9$
A real number squared is never negative, so nothing works
No real solutions.
Divide both sides by 2: $(x - 3)^2 = 25$
Take the square root of both sides: $x - 3 = \pm 5$
Add 3 to both sides: $x = 3 \pm 5$
Split into two cases: $x = 3 + 5 \text{ or } x = 3 - 5$
Simplify: $x = 8 \text{ or } x = -2$
The area of a square is side times side, so let $s$ be the side: $s^2 = 75$
Take the square root of both sides: $s = \pm\sqrt{75}$
The largest perfect square that divides 75 is 25: $s = \pm\sqrt{25 \cdot 3}$
Simplify: $s = \pm 5\sqrt{3}$
A length can’t be negative, so use the positive answer
Each side is $5\sqrt{3}$ feet.

FAQ

Why do you use plus or minus when taking a square root?

Because a positive and a negative number give the same square. Both $7^2$ and $(-7)^2$ equal 49, so $x^2 = 49$ has two solutions.

When can you solve a quadratic with square roots?

When there is no x-term, like $ax^2 + c = 0$ , or when the square is already isolated, like $(x - h)^2 = k$ .

What if the number under the square root is negative?

Then there are no real solutions. For example, $x^2 = -4$ has no real answer. In Algebra 2 these lead to complex numbers.

Do I need to simplify the square root?

Yes. Final answers should be in simplest radical form, like $\pm 2\sqrt{3}$ instead of $\pm\sqrt{12}$ .

Summary

Isolate $x^2$ or $(x - h)^2$ first.
Take the square root of both sides and write $\pm$ .
Simplify the radical, then solve any small equations.
If $x^2$ equals a negative number, there are no real solutions.
In word problems, drop answers like negative time or length.

Previous lesson: Solving Quadratic Equations by Factoring

Next lesson: Completing the Square

Back to the Algebra 1 Index