Solve sqrt{x^2-5}=x-1

$$\sqrt { x ^ { 2 } - 5 } = x - 1$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}-5}\right)^{2}=\left(x-1\right)^{2}$$

Calculate \sqrt{x^{2}-5} to the power of 2 and get x^{2}-5.

$$x^{2}-5=\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-5=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$x^{2}-5-x^{2}=-2x+1$$

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

$$-5=-2x+1$$

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

$$-2x+1=-5$$

Subtract 1 from both sides.

$$-2x=-5-1$$

Subtract 1 from -5 to get -6.

$$-2x=-6$$

Divide both sides by -2.

$$x=\frac{-6}{-2}$$

Divide -6 by -2 to get 3.

$$x=3$$

Substitute 3 for x in the equation \sqrt{x^{2}-5}=x-1.

$$\sqrt{3^{2}-5}=3-1$$

Simplify. The value x=3 satisfies the equation.

$$2=2$$

Equation \sqrt{x^{2}-5}=x-1 has a unique solution.

$$x=3$$