Solve sqrt{x^2-x-3}=x+2

$$\sqrt { x ^ { 2 } - x - 3 } = x + 2$$

Solve for x

$$x = -\frac{7}{5} = -1\frac{2}{5} = -1.4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}-x-3}\right)^{2}=\left(x+2\right)^{2}$$

Calculate \sqrt{x^{2}-x-3} to the power of 2 and get x^{2}-x-3.

$$x^{2}-x-3=\left(x+2\right)^{2}$$

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

$$x^{2}-x-3=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$x^{2}-x-3-x^{2}=4x+4$$

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

$$-x-3=4x+4$$

Subtract 4x from both sides.

$$-x-3-4x=4$$

Combine -x and -4x to get -5x.

$$-5x-3=4$$

Add 3 to both sides.

$$-5x=4+3$$

Add 4 and 3 to get 7.

$$-5x=7$$

Divide both sides by -5.

$$x=\frac{7}{-5}$$

Fraction \frac{7}{-5} can be rewritten as -\frac{7}{5} by extracting the negative sign.

$$x=-\frac{7}{5}$$

Substitute -\frac{7}{5} for x in the equation \sqrt{x^{2}-x-3}=x+2.

$$\sqrt{\left(-\frac{7}{5}\right)^{2}-\left(-\frac{7}{5}\right)-3}=-\frac{7}{5}+2$$

Simplify. The value x=-\frac{7}{5} satisfies the equation.

$$\frac{3}{5}=\frac{3}{5}$$

Equation \sqrt{x^{2}-x-3}=x+2 has a unique solution.

$$x=-\frac{7}{5}$$