Solve for x square root of 2x+5=x-5

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

Step 1

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 2

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

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.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

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

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 3

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 4

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

$x = 10$

Solve for x square root of 2x+5=x-5

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