Solve by Factoring square root of 9x+55-5=x

$\sqrt{9 x + 55} - 5 = x$

Step 1

Subtract $x$ from both sides of the equation.

$\sqrt{9 x + 55} - 5 - x = 0$

Step 2

Move all terms not containing $\sqrt{9 x + 55}$ to the right side of the equation.

Add $5$ to both sides of the equation.

$\sqrt{9 x + 55} - x = 5$

Add $x$ to both sides of the equation.

$\sqrt{9 x + 55} = 5 + x$

Step 3

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

${\sqrt{9 x + 55}}^{2} = {(5 + x)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{9 x + 55}$ as ${(9 x + 55)}^{\frac{1}{2}}$ .

${({(9 x + 55)}^{\frac{1}{2}})}^{2} = {(5 + x)}^{2}$

Simplify the left side.

Simplify ${({(9 x + 55)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(9 x + 55)}^{\frac{1}{2}})}^{2}$ .

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

${(9 x + 55)}^{\frac{1}{2} \cdot 2} = {(5 + x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(9 x + 55)}^{\frac{1}{2} \cdot 2} = {(5 + x)}^{2}$

Rewrite the expression.

${(9 x + 55)}^{1} = {(5 + x)}^{2}$

Simplify.

$9 x + 55 = {(5 + x)}^{2}$

Simplify the right side.

Simplify ${(5 + x)}^{2}$ .

Rewrite ${(5 + x)}^{2}$ as $(5 + x) (5 + x)$ .

$9 x + 55 = (5 + x) (5 + x)$

Expand $(5 + x) (5 + x)$ using the FOIL Method.

Apply the distributive property.

$9 x + 55 = 5 (5 + x) + x (5 + x)$

Apply the distributive property.

$9 x + 55 = 5 \cdot 5 + 5 x + x (5 + x)$

Apply the distributive property.

$9 x + 55 = 5 \cdot 5 + 5 x + x \cdot 5 + x \cdot x$

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$9 x + 55 = 25 + 5 x + x \cdot 5 + x \cdot x$

Move $5$ to the left of $x$ .

$9 x + 55 = 25 + 5 x + 5 \cdot x + x \cdot x$

Multiply $x$ by $x$ .

$9 x + 55 = 25 + 5 x + 5 x + x^{2}$

Add $5 x$ and $5 x$ .

$9 x + 55 = 25 + 10 x + x^{2}$

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.

$25 + 10 x + x^{2} = 9 x + 55$

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

Subtract $9 x$ from both sides of the equation.

$25 + 10 x + x^{2} - 9 x = 55$

Subtract $9 x$ from $10 x$ .

$25 + x^{2} + x = 55$

Subtract $55$ from both sides of the equation.

$25 + x^{2} + x - 55 = 0$

Subtract $55$ from $25$ .

$x^{2} + x - 30 = 0$

Factor $x^{2} + x - 30$ 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 $- 30$ and whose sum is $1$ .

$- 5, 6$

Write the factored form using these integers.

$(x - 5) (x + 6) = 0$

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

$x - 5 = 0$

$x + 6 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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 $(x - 5) (x + 6) = 0$ true.

$x = 5, - 6$

Step 6

Exclude the solutions that do not make $\sqrt{9 x + 55} - 5 - x = 0$ true.

$x = 5$

Solve by Factoring square root of 9x+55-5=x

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