Solve for x x=5+ square root of 2x-10

$x = 5 + \sqrt{2 x - 10}$

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$5 + \sqrt{2 x - 10} = x$

Subtract $5$ from both sides of the equation.

$\sqrt{2 x - 10} = x - 5$

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

${\sqrt{2 x - 10}}^{2} = {(x - 5)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 10}$ as ${(2 x - 10)}^{\frac{1}{2}}$ .

${({(2 x - 10)}^{\frac{1}{2}})}^{2} = {(x - 5)}^{2}$

Simplify the left side.

Simplify ${({(2 x - 10)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 x - 10)}^{\frac{1}{2}})}^{2}$ .

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

${(2 x - 10)}^{\frac{1}{2} \cdot 2} = {(x - 5)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x - 10)}^{\frac{1}{2} \cdot 2} = {(x - 5)}^{2}$

Rewrite the expression.

${(2 x - 10)}^{1} = {(x - 5)}^{2}$

Simplify.

$2 x - 10 = {(x - 5)}^{2}$

Simplify the right side.

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

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

$2 x - 10 = (x - 5) (x - 5)$

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

Apply the distributive property.

$2 x - 10 = x (x - 5) - 5 (x - 5)$

Apply the distributive property.

$2 x - 10 = x \cdot x + x \cdot - 5 - 5 (x - 5)$

Apply the distributive property.

$2 x - 10 = x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$2 x - 10 = x^{2} + x \cdot - 5 - 5 x - 5 \cdot - 5$

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

$2 x - 10 = x^{2} - 5 \cdot x - 5 x - 5 \cdot - 5$

Multiply $- 5$ by $- 5$ .

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

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

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

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.

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

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

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

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

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

$x^{2} - 12 x + 25 = - 10$

Add $10$ to both sides of the equation.

$x^{2} - 12 x + 25 + 10 = 0$

Add $25$ and $10$ .

$x^{2} - 12 x + 35 = 0$

Factor $x^{2} - 12 x + 35$ 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 $35$ and whose sum is $- 12$ .

$- 7, - 5$

Write the factored form using these integers.

$(x - 7) (x - 5) = 0$

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

$x - 7 = 0$

$x - 5 = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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.

$x = 5$

The final solution is all the values that make $(x - 7) (x - 5) = 0$ true.

$x = 7, 5$

Solve for x x=5+ square root of 2x-10

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