Solve by Factoring square root of -3x-5-3=x

$\sqrt{- 3 x - 5} - 3 = x$

Step 1

Subtract $x$ from both sides of the equation.

$\sqrt{- 3 x - 5} - 3 - x = 0$

Step 2

Move all terms not containing $\sqrt{- 3 x - 5}$ to the right side of the equation.

Add $3$ to both sides of the equation.

$\sqrt{- 3 x - 5} - x = 3$

Add $x$ to both sides of the equation.

$\sqrt{- 3 x - 5} = 3 + x$

Step 3

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

${\sqrt{- 3 x - 5}}^{2} = {(3 + x)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- 3 x - 5}$ as ${(- 3 x - 5)}^{\frac{1}{2}}$ .

${({(- 3 x - 5)}^{\frac{1}{2}})}^{2} = {(3 + x)}^{2}$

Simplify the left side.

Simplify ${({(- 3 x - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(- 3 x - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(- 3 x - 5)}^{\frac{1}{2} \cdot 2} = {(3 + x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(- 3 x - 5)}^{\frac{1}{2} \cdot 2} = {(3 + x)}^{2}$

Rewrite the expression.

${(- 3 x - 5)}^{1} = {(3 + x)}^{2}$

Simplify.

$- 3 x - 5 = {(3 + x)}^{2}$

Simplify the right side.

Simplify ${(3 + x)}^{2}$ .

Rewrite ${(3 + x)}^{2}$ as $(3 + x) (3 + x)$ .

$- 3 x - 5 = (3 + x) (3 + x)$

Expand $(3 + x) (3 + x)$ using the FOIL Method.

Apply the distributive property.

$- 3 x - 5 = 3 (3 + x) + x (3 + x)$

Apply the distributive property.

$- 3 x - 5 = 3 \cdot 3 + 3 x + x (3 + x)$

Apply the distributive property.

$- 3 x - 5 = 3 \cdot 3 + 3 x + x \cdot 3 + x \cdot x$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$- 3 x - 5 = 9 + 3 x + x \cdot 3 + x \cdot x$

Move $3$ to the left of $x$ .

$- 3 x - 5 = 9 + 3 x + 3 \cdot x + x \cdot x$

Multiply $x$ by $x$ .

$- 3 x - 5 = 9 + 3 x + 3 x + x^{2}$

Add $3 x$ and $3 x$ .

$- 3 x - 5 = 9 + 6 x + x^{2}$

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.

$9 + 6 x + x^{2} = - 3 x - 5$

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

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

$9 + 6 x + x^{2} + 3 x = - 5$

Add $6 x$ and $3 x$ .

$9 + x^{2} + 9 x = - 5$

Add $5$ to both sides of the equation.

$9 + x^{2} + 9 x + 5 = 0$

Add $9$ and $5$ .

$x^{2} + 9 x + 14 = 0$

Factor $x^{2} + 9 x + 14$ 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 $14$ and whose sum is $9$ .

$2, 7$

Write the factored form using these integers.

$(x + 2) (x + 7) = 0$

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

$x + 2 = 0$

$x + 7 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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 $(x + 2) (x + 7) = 0$ true.

$x = - 2, - 7$

Step 6

Exclude the solutions that do not make $\sqrt{- 3 x - 5} - 3 - x = 0$ true.

$x = - 2$

Solve by Factoring square root of -3x-5-3=x

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