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

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

Solve for x

$$x=\frac{-\sqrt{17}-1}{2}\approx -2.561552813$$

$$x = \frac{\sqrt{17} - 1}{2} \approx 1.561552813$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x from both sides.

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

Combine 2x and -x to get x.

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

Subtract 5 from both sides.

$$x^{2}+x+1-5=0$$

Subtract 5 from 1 to get -4.

$$x^{2}+x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-1±\sqrt{1+16}}{2}$$

Add 1 to 16.

$$x=\frac{-1±\sqrt{17}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{2} when ± is plus. Add -1 to \sqrt{17}.

$$x=\frac{\sqrt{17}-1}{2}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from -1.

$$x=\frac{-\sqrt{17}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-1}{2} x=\frac{-\sqrt{17}-1}{2}$$

Substitute \frac{\sqrt{17}-1}{2} for x in the equation \sqrt{x^{2}+2x+1}=\sqrt{x+5}.

$$\sqrt{\left(\frac{\sqrt{17}-1}{2}\right)^{2}+2\times \frac{\sqrt{17}-1}{2}+1}=\sqrt{\frac{\sqrt{17}-1}{2}+5}$$

Simplify. The value x=\frac{\sqrt{17}-1}{2} satisfies the equation.

$$\frac{1}{2}\times 17^{\frac{1}{2}}+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}\times 17^{\frac{1}{2}}$$

Substitute \frac{-\sqrt{17}-1}{2} for x in the equation \sqrt{x^{2}+2x+1}=\sqrt{x+5}.

$$\sqrt{\left(\frac{-\sqrt{17}-1}{2}\right)^{2}+2\times \frac{-\sqrt{17}-1}{2}+1}=\sqrt{\frac{-\sqrt{17}-1}{2}+5}$$

Simplify. The value x=\frac{-\sqrt{17}-1}{2} satisfies the equation.

$$\frac{1}{2}\times 17^{\frac{1}{2}}-\frac{1}{2}=-\left(\frac{1}{2}-\frac{1}{2}\times 17^{\frac{1}{2}}\right)$$

List all solutions of \sqrt{x^{2}+2x+1}=\sqrt{x+5}.

$$x=\frac{\sqrt{17}-1}{2} x=\frac{-\sqrt{17}-1}{2}$$