Solve sqrt{x^2-1}=sqrt{2x+2}

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

Solve for x

$$x=3$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}-1} to the power of 2 and get x^{2}-1.

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

Calculate \sqrt{2x+2} to the power of 2 and get 2x+2.

$$x^{2}-1=2x+2$$

Subtract 2x from both sides.

$$x^{2}-1-2x=2$$

Subtract 2 from both sides.

$$x^{2}-1-2x-2=0$$

Subtract 2 from -1 to get -3.

$$x^{2}-3-2x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-2x-3=0$$

To solve the equation, factor x^{2}-2x-3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+1\right)$$

To find equation solutions, solve x-3=0 and x+1=0.

$$x=3 x=-1$$

Substitute 3 for x in the equation \sqrt{x^{2}-1}=\sqrt{2x+2}.

$$\sqrt{3^{2}-1}=\sqrt{2\times 3+2}$$

Simplify. The value x=3 satisfies the equation.

$$2\times 2^{\frac{1}{2}}=2\times 2^{\frac{1}{2}}$$

Substitute -1 for x in the equation \sqrt{x^{2}-1}=\sqrt{2x+2}.

$$\sqrt{\left(-1\right)^{2}-1}=\sqrt{2\left(-1\right)+2}$$

Simplify. The value x=-1 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{x^{2}-1}=\sqrt{2x+2}.

$$x=3 x=-1$$