Solve for x square root of -3x-2=x+2

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$- 3 x - 2 = (x + 2) (x + 2)$

Expand $(x + 2) (x + 2)$ using the FOIL Method.

Apply the distributive property.

$- 3 x - 2 = x (x + 2) + 2 (x + 2)$

Apply the distributive property.

$- 3 x - 2 = x \cdot x + x \cdot 2 + 2 (x + 2)$

Apply the distributive property.

$- 3 x - 2 = x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $2$ to the left of $x$ .

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

Multiply $2$ by $2$ .

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

Add $2 x$ and $2 x$ .

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

Step 3

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.

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

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

Add $3 x$ to both sides of the equation.

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

Add $4 x$ and $3 x$ .

$x^{2} + 7 x + 4 = - 2$

Add $2$ to both sides of the equation.

$x^{2} + 7 x + 4 + 2 = 0$

Add $4$ and $2$ .

$x^{2} + 7 x + 6 = 0$

Factor $x^{2} + 7 x + 6$ 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 $6$ and whose sum is $7$ .

$1, 6$

Write the factored form using these integers.

$(x + 1) (x + 6) = 0$

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

$x + 1 = 0$

$x + 6 = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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 $(x + 1) (x + 6) = 0$ true.

$x = - 1, - 6$

Step 4

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

$x = - 1$

Solve for x square root of -3x-2=x+2

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