Evaluate square root of -2x-5-4=x

$\sqrt{- 2 x - 5} - 4 = x$

Step 1

Add $4$ to both sides of the equation.

$\sqrt{- 2 x - 5} = x + 4$

Step 2

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

${\sqrt{- 2 x - 5}}^{2} = {(x + 4)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- 2 x - 5}$ as ${(- 2 x - 5)}^{\frac{1}{2}}$ .

${({(- 2 x - 5)}^{\frac{1}{2}})}^{2} = {(x + 4)}^{2}$

Simplify the left side.

Simplify ${({(- 2 x - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(- 2 x - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(- 2 x - 5)}^{\frac{1}{2} \cdot 2} = {(x + 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(- 2 x - 5)}^{\frac{1}{2} \cdot 2} = {(x + 4)}^{2}$

Rewrite the expression.

${(- 2 x - 5)}^{1} = {(x + 4)}^{2}$

Simplify.

$- 2 x - 5 = {(x + 4)}^{2}$

Simplify the right side.

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

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

$- 2 x - 5 = (x + 4) (x + 4)$

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

Apply the distributive property.

$- 2 x - 5 = x (x + 4) + 4 (x + 4)$

Apply the distributive property.

$- 2 x - 5 = x \cdot x + x \cdot 4 + 4 (x + 4)$

Apply the distributive property.

$- 2 x - 5 = x \cdot x + x \cdot 4 + 4 x + 4 \cdot 4$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$- 2 x - 5 = x^{2} + x \cdot 4 + 4 x + 4 \cdot 4$

Move $4$ to the left of $x$ .

$- 2 x - 5 = x^{2} + 4 \cdot x + 4 x + 4 \cdot 4$

Multiply $4$ by $4$ .

$- 2 x - 5 = x^{2} + 4 x + 4 x + 16$

Add $4 x$ and $4 x$ .

$- 2 x - 5 = x^{2} + 8 x + 16$

Step 4

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.

$x^{2} + 8 x + 16 = - 2 x - 5$

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

Add $2 x$ to both sides of the equation.

$x^{2} + 8 x + 16 + 2 x = - 5$

Add $8 x$ and $2 x$ .

$x^{2} + 10 x + 16 = - 5$

Add $5$ to both sides of the equation.

$x^{2} + 10 x + 16 + 5 = 0$

Add $16$ and $5$ .

$x^{2} + 10 x + 21 = 0$

Factor $x^{2} + 10 x + 21$ 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 $21$ and whose sum is $10$ .

$3, 7$

Write the factored form using these integers.

$(x + 3) (x + 7) = 0$

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

$x + 3 = 0$

$x + 7 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x + 3) (x + 7) = 0$ true.

$x = - 3, - 7$

Step 5

Exclude the solutions that do not make $\sqrt{- 2 x - 5} - 4 = x$ true.

$x = - 3$

Evaluate square root of -2x-5-4=x

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