Allday Education

Solving Quadratics by Square Roots

When a quadratic has no middle xx term, like x2=81x^2 = 81, you do not need factoring or the formula. Just isolate the squared part and take the square root.

The one thing you must not forget is the plus-or-minus. Every positive number has two square roots, so a quadratic solved this way usually has two answers.

Isolate, then root

Get the squared term alone on one side. Then take the square root of both sides, writing ±\pm in front of the root.

For x2=81x^2 = 81, the square root gives x=±9x = \pm 9. Both 99 and −9-9 square to 8181, so both are solutions.

When the inside is a binomial

If the squared part is something like (x−3)2=16(x - 3)^2 = 16, take the root to get x−3=±4x - 3 = \pm 4, then solve the two resulting equations.

A negative on the right, like x2=−25x^2 = -25, gives imaginary solutions x=±5ix = \pm 5i — no real answers.

Worked examples

Example 1: a basic square-root solve

Solve x2=81x^2 = 81.

Take the square root of both sidesx=±81x = \pm\sqrt{81}
Simplifyx=±9x = \pm 9

Answer: x=9x = 9 or x=−9x = -9

Example 2: a squared binomial

Solve (x−3)2=16(x - 3)^2 = 16.

Take the square rootx−3=±4x - 3 = \pm 4
Solve bothx=7 or x=−1x = 7 \text{ or } x = -1

Answer: x=7x = 7 or x=−1x = -1

Example 3: a negative on the right side

Solve x2=−25x^2 = -25.

Take the square root of both sidesx=±−25x = \pm\sqrt{-25}
The square root of a negative is imaginaryx=±5ix = \pm 5i

Answer: x=5ix = 5i or x=−5ix = -5i, with no real solutions

Try one yourself

Common questions

Why the plus-or-minus?

Both a positive and a negative number square to the same positive result, so a squared quantity has two square roots. Missing the ±\pm loses half the solutions.

When can I use this method?

When the quadratic has no linear (middle) term, or when one side is already a perfect square like (x−3)2(x - 3)^2.

What if the right side is negative?

The solutions are imaginary. x2=−25x^2 = -25 gives x=±5ix = \pm 5i — there are no real solutions.

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