Solve by Factoring x- square root of 2x+5=5

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

Step 1

Subtract $5$ from both sides of the equation.

$x - \sqrt{2 x + 5} - 5 = 0$

Step 2

Move all terms not containing $- \sqrt{2 x + 5}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

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

Add $5$ to both sides of the equation.

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

Step 3

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

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

Step 4

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $- {(2 x + 5)}^{\frac{1}{2}}$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply ${({(2 x + 5)}^{\frac{1}{2}})}^{2}$ by $1$ .

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 x + 5 = - 1 \cdot - 1 x \cdot x - x \cdot 5 + 5 (- x) + 5 \cdot 5$

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

Move $x$ .

$2 x + 5 = - 1 \cdot - 1 (x \cdot x) - x \cdot 5 + 5 (- x) + 5 \cdot 5$

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Multiply $5$ by $- 1$ .

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

Multiply $- 1$ by $5$ .

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

Multiply $5$ by $5$ .

$2 x + 5 = x^{2} - 5 x - 5 x + 25$

Subtract $5 x$ from $- 5 x$ .

$2 x + 5 = x^{2} - 10 x + 25$

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.

$x^{2} - 10 x + 25 = 2 x + 5$

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

Subtract $2 x$ from both sides of the equation.

$x^{2} - 10 x + 25 - 2 x = 5$

Subtract $2 x$ from $- 10 x$ .

$x^{2} - 12 x + 25 = 5$

Subtract $5$ from both sides of the equation.

$x^{2} - 12 x + 25 - 5 = 0$

Subtract $5$ from $25$ .

$x^{2} - 12 x + 20 = 0$

Factor $x^{2} - 12 x + 20$ 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 $20$ and whose sum is $- 12$ .

$- 10, - 2$

Write the factored form using these integers.

$(x - 10) (x - 2) = 0$

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

$x - 10 = 0$

$x - 2 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x - 10) (x - 2) = 0$ true.

$x = 10, 2$

Step 6

Exclude the solutions that do not make $x - \sqrt{2 x + 5} - 5 = 0$ true.

$x = 10$

Solve by Factoring x- square root of 2x+5=5

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