Solve for x square root of x^2+2x-25 = square root of x+5

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

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

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

Simplify each side of the equation.

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Multiply the exponents in ${({(x^{2} + 2 x - 25)}^{\frac{1}{2}})}^{2}$ .

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

${(x^{2} + 2 x - 25)}^{\frac{1}{2} \cdot 2} = {\sqrt{x + 5}}^{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

${(x^{2} + 2 x - 25)}^{\frac{1}{2} \cdot 2} = {\sqrt{x + 5}}^{2}$

Rewrite the expression.

${(x^{2} + 2 x - 25)}^{1} = {\sqrt{x + 5}}^{2}$

Simplify.

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

Rewrite ${\sqrt{x + 5}}^{2}$ as $x + 5$ .

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

$x^{2} + 2 x - 25 = {({(x + 5)}^{\frac{1}{2}})}^{2}$

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

$x^{2} + 2 x - 25 = {(x + 5)}^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$x^{2} + 2 x - 25 = {(x + 5)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$x^{2} + 2 x - 25 = {(x + 5)}^{\frac{2}{2}}$

Divide $1$ by $1$ .

$x^{2} + 2 x - 25 = {(x + 5)}^{1}$

Simplify.

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

Solve for $x$ .

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

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Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $2 x$ .

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

Move $5$ to the left side of the equation by subtracting it from both sides.

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

Subtract $5$ from $- 25$ .

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

Factor $x^{2} + x - 30$ using the AC method.

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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 the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set the next factor equal to $0$ and solve.

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Set the next factor 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.

$x = 5, - 6$

Exclude the solutions that do not make $\sqrt{x^{2} + 2 x - 25} = \sqrt{x + 5}$ true.

$x = 5$

Solve for x square root of x^2+2x-25 = square root of x+5