# Solving Radical Equations

How to solve radical equations (square root equations): example and its solution.

## Example: Solve √*x* + 6 = *x*

Before solving the radical equation,

find the range of *x*

from the given equation.

For an even root (√, ^{4}√, ^{6}√, etc.),

the radicand (the number inside the radical)

cannot be (-).

So the radicand of √*x* + 6, [*x* + 6],

cannot be (-).

So *x* + 6 ≥ 0.

Move +6 to the right side.

Then *x* ≥ -6.

The left side, √*x* + 6, is not (-).

There's no (-) sign.

So the right side, *x*, is also not (-).

So *x* ≥ 0.

*x* ≥ -6*x* ≥ 0

Draw these two inequalities on a number line.

And find the intersection.

Then *x* ≥ 0.

Next, solve the radical equation.

Square both sides.

(√*x* + 6)^{2} = *x* + 6

Square root - Square of a square root

Move *x*^{2} to the left side.

Then -*x*^{2} + *x* + 6 = 0.

Multiply -1 on both sides.

Then *x*^{2} - *x* - 6 = 0.

Factor *x*^{2} - *x* - 6.

Factor a quadratic trinomial

Find a pair of numbers

whose product is the constant term [-6]

and whose sum is the middle term's coefficient [-1].

The constant term is (-).

So the signs of the numbers are different:

one is (+), and the other is (-).

(-1, 6) and (-2, 3)

are not the right numbers.

[-6] = -3⋅2

-3 + 2 = [-1]

So -3 and 2 are the right numbers.

Use -3 and +2

to write a factored form:

(*x* - 3)(*x* + 2) = 0.

Solve (*x* - 3)(*x* + 2) = 0.

1) *x* - 3 = 0

So *x* = 3.

2) *x* + 2 = 0

So *x* = -2.

Solving a quadratic equation by factoring

See if the found *x* values

are in the initial range of *x*:*x* ≥ 0.*x* = 3 is in the range.

But *x* = -2 is not in the range.

So *x* = 3 is the answer.