Solve for x square root of x^2-2 = square root of x+4

$\sqrt{x^{2} - 2} = \sqrt{x + 4}$

Step 1

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

${\sqrt{x^{2} - 2}}^{2} = {\sqrt{x + 4}}^{2}$

Step 2

Simplify each side of the equation.

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

${({(x^{2} - 2)}^{\frac{1}{2}})}^{2} = {\sqrt{x + 4}}^{2}$

Simplify the left side.

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

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

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

${(x^{2} - 2)}^{\frac{1}{2} \cdot 2} = {\sqrt{x + 4}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} - 2)}^{\frac{1}{2} \cdot 2} = {\sqrt{x + 4}}^{2}$

Rewrite the expression.

${(x^{2} - 2)}^{1} = {\sqrt{x + 4}}^{2}$

Simplify.

$x^{2} - 2 = {\sqrt{x + 4}}^{2}$

Simplify the right side.

Rewrite ${\sqrt{x + 4}}^{2}$ as $x + 4$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 4}$ as ${(x + 4)}^{\frac{1}{2}}$ .

$x^{2} - 2 = {({(x + 4)}^{\frac{1}{2}})}^{2}$

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

$x^{2} - 2 = {(x + 4)}^{\frac{1}{2} \cdot 2}$

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

$x^{2} - 2 = {(x + 4)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{2} - 2 = {(x + 4)}^{\frac{2}{2}}$

Rewrite the expression.

$x^{2} - 2 = {(x + 4)}^{1}$

Simplify.

$x^{2} - 2 = x + 4$

Step 3

Solve for $x$ .

Subtract $x$ from both sides of the equation.

$x^{2} - 2 - x = 4$

Subtract $4$ from both sides of the equation.

$x^{2} - 2 - x - 4 = 0$

Subtract $4$ from $- 2$ .

$x^{2} - x - 6 = 0$

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$(x - 3) (x + 2) = 0$

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

$x - 3 = 0$

$x + 2 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Solve for x square root of x^2-2 = square root of x+4

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