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

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

Solve for x

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Solution Steps

Subtract x from both sides of the equation.

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

Square both sides of the equation.

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

Expand \left(-\sqrt{x^{2}-5}\right)^{2}.

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

Calculate -1 to the power of 2 and get 1.

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

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

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

Use the distributive property to multiply 1 by x^{2}-5.

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

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

$$x^{2}-5=4-4x+x^{2}$$

Add 4x to both sides.

$$x^{2}-5+4x=4+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-5+4x-x^{2}=4$$

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

$$-5+4x=4$$

Add 5 to both sides.

$$4x=4+5$$

Add 4 and 5 to get 9.

$$4x=9$$

Divide both sides by 4.

$$x=\frac{9}{4}$$

Substitute \frac{9}{4} for x in the equation x-\sqrt{x^{2}-5}=2.

$$\frac{9}{4}-\sqrt{\left(\frac{9}{4}\right)^{2}-5}=2$$

Simplify. The value x=\frac{9}{4} satisfies the equation.

$$2=2$$

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

$$x=\frac{9}{4}$$