Solve for x 6+3x = square root of 2x+12+2x

$6 + 3 x = \sqrt{2 x + 12} + 2 x$

Step 1

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

$\sqrt{2 x + 12} + 2 x = 6 + 3 x$

Step 2

Move all terms not containing $\sqrt{2 x + 12}$ to the right side of the equation.

Subtract $2 x$ from both sides of the equation.

$\sqrt{2 x + 12} = 6 + 3 x - 2 x$

Subtract $2 x$ from $3 x$ .

$\sqrt{2 x + 12} = 6 + x$

Step 3

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

${\sqrt{2 x + 12}}^{2} = {(6 + x)}^{2}$

Step 4

Simplify each side of the equation.

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

${({(2 x + 12)}^{\frac{1}{2}})}^{2} = {(6 + x)}^{2}$

Simplify the left side.

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

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

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

${(2 x + 12)}^{\frac{1}{2} \cdot 2} = {(6 + x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x + 12)}^{\frac{1}{2} \cdot 2} = {(6 + x)}^{2}$

Rewrite the expression.

${(2 x + 12)}^{1} = {(6 + x)}^{2}$

Simplify.

$2 x + 12 = {(6 + x)}^{2}$

Simplify the right side.

Simplify ${(6 + x)}^{2}$ .

Rewrite ${(6 + x)}^{2}$ as $(6 + x) (6 + x)$ .

$2 x + 12 = (6 + x) (6 + x)$

Expand $(6 + x) (6 + x)$ using the FOIL Method.

Apply the distributive property.

$2 x + 12 = 6 (6 + x) + x (6 + x)$

Apply the distributive property.

$2 x + 12 = 6 \cdot 6 + 6 x + x (6 + x)$

Apply the distributive property.

$2 x + 12 = 6 \cdot 6 + 6 x + x \cdot 6 + x \cdot x$

Simplify and combine like terms.

Simplify each term.

Multiply $6$ by $6$ .

$2 x + 12 = 36 + 6 x + x \cdot 6 + x \cdot x$

Move $6$ to the left of $x$ .

$2 x + 12 = 36 + 6 x + 6 \cdot x + x \cdot x$

Multiply $x$ by $x$ .

$2 x + 12 = 36 + 6 x + 6 x + x^{2}$

Add $6 x$ and $6 x$ .

$2 x + 12 = 36 + 12 x + x^{2}$

Step 5

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.

$36 + 12 x + x^{2} = 2 x + 12$

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

Subtract $2 x$ from both sides of the equation.

$36 + 12 x + x^{2} - 2 x = 12$

Subtract $2 x$ from $12 x$ .

$36 + x^{2} + 10 x = 12$

Subtract $12$ from both sides of the equation.

$36 + x^{2} + 10 x - 12 = 0$

Subtract $12$ from $36$ .

$x^{2} + 10 x + 24 = 0$

Factor $x^{2} + 10 x + 24$ 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 $24$ and whose sum is $10$ .

$4, 6$

Write the factored form using these integers.

$(x + 4) (x + 6) = 0$

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

$x + 4 = 0$

$x + 6 = 0$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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 $(x + 4) (x + 6) = 0$ true.

$x = - 4, - 6$

Solve for x 6+3x = square root of 2x+12+2x

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