Solve by Factoring x = square root of 3x-12+4

$x = \sqrt{3 x - 12} + 4$

Step 1

Move all the expressions to the left side of the equation.

Subtract $\sqrt{3 x - 12}$ from both sides of the equation.

$x - \sqrt{3 x - 12} = 4$

Subtract $4$ from both sides of the equation.

$x - \sqrt{3 x - 12} - 4 = 0$

Step 2

Factor $3$ out of $3 x - 12$ .

Factor $3$ out of $3 x$ .

$x - \sqrt{3 (x) - 12} - 4 = 0$

Factor $3$ out of $- 12$ .

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

Factor $3$ out of $3 x + 3 \cdot - 4$ .

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

Step 3

Move all terms not containing $- \sqrt{3 (x - 4)}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

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

Add $4$ to both sides of the equation.

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

Step 4

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

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

Step 5

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 (x - 4)}$ as ${(3 (x - 4))}^{\frac{1}{2}}$ .

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

Simplify the left side.

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

Apply the distributive property.

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

Simplify the expression.

Multiply $3$ by $- 4$ .

${(- {(3 x - 12)}^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

Apply the product rule to $- {(3 x - 12)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(3 x - 12)}^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(3 x - 12)}^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

Multiply ${({(3 x - 12)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(3 x - 12)}^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

Multiply the exponents in ${({(3 x - 12)}^{\frac{1}{2}})}^{2}$ .

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

${(3 x - 12)}^{\frac{1}{2} \cdot 2} = {(- x + 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x - 12)}^{\frac{1}{2} \cdot 2} = {(- x + 4)}^{2}$

Rewrite the expression.

${(3 x - 12)}^{1} = {(- x + 4)}^{2}$

Simplify.

$3 x - 12 = {(- x + 4)}^{2}$

Simplify the right side.

Simplify ${(- x + 4)}^{2}$ .

Rewrite ${(- x + 4)}^{2}$ as $(- x + 4) (- x + 4)$ .

$3 x - 12 = (- x + 4) (- x + 4)$

Expand $(- x + 4) (- x + 4)$ using the FOIL Method.

Apply the distributive property.

$3 x - 12 = - x (- x + 4) + 4 (- x + 4)$

Apply the distributive property.

$3 x - 12 = - x (- x) - x \cdot 4 + 4 (- x + 4)$

Apply the distributive property.

$3 x - 12 = - x (- x) - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 x - 12 = - 1 \cdot - 1 x \cdot x - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$3 x - 12 = - 1 \cdot - 1 (x \cdot x) - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $x$ by $x$ .

$3 x - 12 = - 1 \cdot - 1 x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $- 1$ by $- 1$ .

$3 x - 12 = 1 x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $x^{2}$ by $1$ .

$3 x - 12 = x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $4$ by $- 1$ .

$3 x - 12 = x^{2} - 4 x + 4 (- x) + 4 \cdot 4$

Multiply $- 1$ by $4$ .

$3 x - 12 = x^{2} - 4 x - 4 x + 4 \cdot 4$

Multiply $4$ by $4$ .

$3 x - 12 = x^{2} - 4 x - 4 x + 16$

Subtract $4 x$ from $- 4 x$ .

$3 x - 12 = x^{2} - 8 x + 16$

Step 6

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 8 x + 16 = 3 x - 12$

Move all terms containing $x$ to the left side of the equation.

Subtract $3 x$ from both sides of the equation.

$x^{2} - 8 x + 16 - 3 x = - 12$

Subtract $3 x$ from $- 8 x$ .

$x^{2} - 11 x + 16 = - 12$

Add $12$ to both sides of the equation.

$x^{2} - 11 x + 16 + 12 = 0$

Add $16$ and $12$ .

$x^{2} - 11 x + 28 = 0$

Factor $x^{2} - 11 x + 28$ 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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

Write the factored form using these integers.

$(x - 7) (x - 4) = 0$

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

$x - 7 = 0$

$x - 4 = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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.

$x = 4$

The final solution is all the values that make $(x - 7) (x - 4) = 0$ true.

$x = 7, 4$

Solve by Factoring x = square root of 3x-12+4

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