Evaluate square root of x+12=x

$\sqrt{x + 12} = x$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x + 12}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 12}$ as ${(x + 12)}^{\frac{1}{2}}$ .

${({(x + 12)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(x + 12)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 12)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x + 12)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 12)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(x + 12)}^{1} = x^{2}$

Simplify.

$x + 12 = x^{2}$

Solve for $x$ .

Subtract $x^{2}$ from both sides of the equation.

$x + 12 - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $x + 12 - x^{2}$ .

Reorder the expression.

Move $12$ .

$x - x^{2} + 12 = 0$

Reorder $x$ and $- x^{2}$ .

$- x^{2} + x + 12 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + x + 12 = 0$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 12 = 0$

Rewrite $12$ as $- 1 (- 12)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 12)$ .

$- (x^{2} - x - 12) = 0$

Factor.

Factor $x^{2} - x - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$- ((x - 4) (x + 3)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x + 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 3 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $- (x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Exclude the solutions that do not make $\sqrt{x + 12} = x$ true.

$x = 4$

Evaluate square root of x+12=x

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