Solve for x square root of 12-4x=x

$\sqrt{12 - 4 x} = x$

Step 1

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

${\sqrt{12 - 4 x}}^{2} = x^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{12 - 4 x}$ as ${(12 - 4 x)}^{\frac{1}{2}}$ .

${({(12 - 4 x)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(12 - 4 x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(12 - 4 x)}^{\frac{1}{2}})}^{2}$ .

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

${(12 - 4 x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(12 - 4 x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(12 - 4 x)}^{1} = x^{2}$

Simplify.

$12 - 4 x = x^{2}$

Step 3

Solve for $x$ .

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

$12 - 4 x - x^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $12 - 4 x - x^{2}$ .

Reorder the expression.

Move $12$ .

$- 4 x - x^{2} + 12 = 0$

Reorder $- 4 x$ and $- x^{2}$ .

$- x^{2} - 4 x + 12 = 0$

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

$- (x^{2}) - 4 x + 12 = 0$

Factor $- 1$ out of $- 4 x$ .

$- (x^{2}) - (4 x) + 12 = 0$

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

$- (x^{2}) - (4 x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2}) - (4 x)$ .

$- (x^{2} + 4 x) - 1 \cdot - 12 = 0$

Factor $- 1$ out of $- (x^{2} + 4 x) - 1 (- 12)$ .

$- (x^{2} + 4 x - 12) = 0$

Factor.

Factor $x^{2} + 4 x - 12$ 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 $4$ .

$- 2, 6$

Write the factored form using these integers.

$- ((x - 2) (x + 6)) = 0$

Remove unnecessary parentheses.

$- (x - 2) (x + 6) = 0$

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

$x - 2 = 0$

$x + 6 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- (x - 2) (x + 6) = 0$ true.

$x = 2, - 6$

Step 4

Exclude the solutions that do not make $\sqrt{12 - 4 x} = x$ true.

$x = 2$

Solve for x square root of 12-4x=x

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