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

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

Solve for x

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

$$-1=4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$4x+4=-1$$

Subtract 4 from both sides.

$$4x=-1-4$$

Subtract 4 from -1 to get -5.

$$4x=-5$$

Divide both sides by 4.

$$x=\frac{-5}{4}$$

Fraction \frac{-5}{4} can be rewritten as -\frac{5}{4} by extracting the negative sign.

$$x=-\frac{5}{4}$$

Substitute -\frac{5}{4} for x in the equation \sqrt{x^{2}-1}=x+2.

$$\sqrt{\left(-\frac{5}{4}\right)^{2}-1}=-\frac{5}{4}+2$$

Simplify. The value x=-\frac{5}{4} satisfies the equation.

$$\frac{3}{4}=\frac{3}{4}$$

Equation \sqrt{x^{2}-1}=x+2 has a unique solution.

$$x=-\frac{5}{4}$$