Square Root Method

Q: Why can't I just write $x = 7$ for $x^{2} = 49$?

Because $-7$ works too: $(-7)^{2} = 49$. The equation has two solutions, and dropping the $\pm$ silently discards one of them.

Q: What if the number is not a perfect square?

Leave the answer as a simplified radical. For $x^{2} = 20$, write $x = \pm\sqrt{20} = \pm 2\sqrt{5}$. A radical is a perfectly good exact answer.

Some quadratic equations have a squared term and a constant but no plain $x$ term: $x^{2} = 49$ , $3x^{2} = 48$ , $(x + 3)^{2} = 16$ . For these, the square root method is the shortest path: isolate the squared part, take the square root of both sides, and write $\pm$ in front of the root.

The whole rule fits in one line: if $x^{2} = k$ , then $x = \pm\sqrt{k}$ . The $\pm$ is not decoration — forgetting it costs you exactly half of your answers, every time.

When to use it

Reach for this method when the equation has no linear term — no plain $x$ floating around. If you see $ax^{2} = k$ or a squared binomial equal to a number, the square root method applies. If there is a $bx$ term, like $x^{2} + 5x + 6 = 0$ , factor instead or use the quadratic formula.

The routine: isolate the squared expression first (divide away any coefficient, move any constant), then take the square root of both sides, then finish solving for $x$ .

Why plus-or-minus

Squaring destroys the sign. Both $7^{2} = 49$ and $(-7)^{2} = 49$ , so when all you know is $x^{2} = 49$ , the $x$ could be either $7$ or $-7$ . Writing $x = \pm 7$ records both possibilities.

This is the single most common lost point on this topic. The square root of $49$ as a number is $7$ , but the solutions of the equation $x^{2} = 49$ are $7$ and $-7$ — the equation question and the square-root question are not the same question.

Squared binomials

The method works even when the squared thing is a whole expression. In $(x + 3)^{2} = 16$ , treat the parentheses as one unit: take the square root of both sides to get $x + 3 = \pm 4$ . That splits into two small equations — $x + 3 = 4$ and $x + 3 = -4$ — giving $x = 1$ or $x = -7$ .

Worked examples

Example 1: the basic case

Solve $x^{2} = 49$ .

Take the square root of both sides — keep the

\pm

x = \pm\sqrt{49}

Simplify

x = \pm 7

Check both:

7^{2} = 49

and

(-7)^{2} = 49

✓

Answer: $x = \pm 7$

Example 2: isolate the square first

Solve $3x^{2} = 48$ .

Divide both sides by

3

to isolate

x^{2}

x^{2} = 16

Take the square root of both sides — keep the

\pm

x = \pm\sqrt{16}

Simplify

x = \pm 4

Answer: $x = \pm 4$

Example 3: a squared binomial

Solve $(x + 3)^{2} = 16$ .

Take the square root of both sides

x + 3 = \pm 4

Split into two equations

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

Solve each

x = 1 \text{ or } x = -7

Answer: $x = 1$ or $x = -7$

Try one yourself

Common questions

Why can't I just write

x = 7

for

x^{2} = 49

Because $-7$ works too: $(-7)^{2} = 49$ . The equation has two solutions, and dropping the $\pm$ silently discards one of them.

What if the number is not a perfect square?

Leave the answer as a simplified radical. For $x^{2} = 20$ , write $x = \pm\sqrt{20} = \pm 2\sqrt{5}$ . A radical is a perfectly good exact answer.

What if

x^{2}

equals a negative number?

Then there are no real solutions. No real number squares to a negative — squaring always gives zero or a positive result. An equation like $x^{2} = -9$ has no real answer in Algebra 1.

How is this different from factoring?

Factoring handles quadratics with a middle term, like $x^{2} + 5x + 6 = 0$ . The square root method skips factoring entirely, but it only works when there is no linear term to deal with.

Want the video version?

Allday Everyday Math has video lessons, practice, and an AI tutor for every topic, Pre-Algebra through Algebra 2.

Try it for $1