Solve by Factoring square root of 12-4x=x

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

Step 1

Subtract $x$ from both sides of the equation.

$\sqrt{12 - 4 x} - x = 0$

Step 2

Simplify each term.

Factor $4$ out of $12 - 4 x$ .

Factor $4$ out of $12$ .

$\sqrt{4 (3) - 4 x} - x = 0$

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

$\sqrt{4 (3) + 4 (- x)} - x = 0$

Factor $4$ out of $4 (3) + 4 (- x)$ .

$\sqrt{4 (3 - x)} - x = 0$

Rewrite $4$ as $2^{2}$ .

$\sqrt{2^{2} (3 - x)} - x = 0$

Pull terms out from under the radical.

$2 \sqrt{3 - x} - x = 0$

Step 3

Add $x$ to both sides of the equation.

$2 \sqrt{3 - x} = x$

Step 4

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

${(2 \sqrt{3 - x})}^{2} = x^{2}$

Step 5

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 - x}$ as ${(3 - x)}^{\frac{1}{2}}$ .

${(2 {(3 - x)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${(2 {(3 - x)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 {(3 - x)}^{\frac{1}{2}}$ .

$2^{2} {({(3 - x)}^{\frac{1}{2}})}^{2} = x^{2}$

Raise $2$ to the power of $2$ .

$4 {({(3 - x)}^{\frac{1}{2}})}^{2} = x^{2}$

Multiply the exponents in ${({(3 - x)}^{\frac{1}{2}})}^{2}$ .

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

$4 {(3 - x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(3 - x)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

$4 {(3 - x)}^{1} = x^{2}$

Simplify.

$4 (3 - x) = x^{2}$

Apply the distributive property.

$4 \cdot 3 + 4 (- x) = x^{2}$

Multiply.

Multiply $4$ by $3$ .

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

Multiply $- 1$ by $4$ .

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

Step 6

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 7

Exclude the solutions that do not make $2 \sqrt{3 - x} - x = 0$ true.

$x = 2$

Solve by Factoring square root of 12-4x=x

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