Solve for x -2 square root of x+3=-x-4

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $4$ by $3$ .

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Multiply $- 4$ by $- 1$ .

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

Multiply $- 1$ by $- 4$ .

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

Multiply $- 4$ by $- 4$ .

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

Add $4 x$ and $4 x$ .

$4 x + 12 = x^{2} + 8 x + 16$

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.

$x^{2} + 8 x + 16 = 4 x + 12$

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

Subtract $4 x$ from both sides of the equation.

$x^{2} + 8 x + 16 - 4 x = 12$

Subtract $4 x$ from $8 x$ .

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

Subtract $12$ from both sides of the equation.

$x^{2} + 4 x + 16 - 12 = 0$

Subtract $12$ from $16$ .

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

Factor using the perfect square rule.

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 2 + 2^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 2$ .

${(x + 2)}^{2} = 0$

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Solve for x -2 square root of x+3=-x-4

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