Solve for r square root of 12-r=r

$\sqrt{12 - r} = r$

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

${\sqrt{12 - r}}^{2} = r^{2}$

Simplify each side of the equation.

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

${({(12 - r)}^{\frac{1}{2}})}^{2} = r^{2}$

Simplify the left side.

Simplify ${({(12 - r)}^{\frac{1}{2}})}^{2}$ .

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

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

${(12 - r)}^{\frac{1}{2} \cdot 2} = r^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(12 - r)}^{\frac{1}{2} \cdot 2} = r^{2}$

Rewrite the expression.

${(12 - r)}^{1} = r^{2}$

Simplify.

$12 - r = r^{2}$

Solve for $r$ .

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

$12 - r - r^{2} = 0$

Factor the left side of the equation.

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

Reorder the expression.

Move $12$ .

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

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

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

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

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

Factor $- 1$ out of $- r$ .

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

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

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

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

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

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

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

Factor.

Factor $r^{2} + r - 12$ 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 $1$ .

$- 3, 4$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$r - 3 = 0$

$r + 4 = 0$

Set $r - 3$ equal to $0$ and solve for $r$ .

Set $r - 3$ equal to $0$ .

$r - 3 = 0$

Add $3$ to both sides of the equation.

$r = 3$

Set $r + 4$ equal to $0$ and solve for $r$ .

Set $r + 4$ equal to $0$ .

$r + 4 = 0$

Subtract $4$ from both sides of the equation.

$r = - 4$

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

$r = 3, - 4$

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

$r = 3$

Solve for r square root of 12-r=r

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