Solve sqrt{16+x^2}=x+1

$$\sqrt { 16 + x ^ { 2 } } = x + 1$$

Solve for x

$$x = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{16+x^{2}}\right)^{2}=\left(x+1\right)^{2}$$

Calculate \sqrt{16+x^{2}} to the power of 2 and get 16+x^{2}.

$$16+x^{2}=\left(x+1\right)^{2}$$

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

$$16+x^{2}=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$16+x^{2}-x^{2}=2x+1$$

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

$$16=2x+1$$

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

$$2x+1=16$$

Subtract 1 from both sides.

$$2x=16-1$$

Subtract 1 from 16 to get 15.

$$2x=15$$

Divide both sides by 2.

$$x=\frac{15}{2}$$

Substitute \frac{15}{2} for x in the equation \sqrt{16+x^{2}}=x+1.

$$\sqrt{16+\left(\frac{15}{2}\right)^{2}}=\frac{15}{2}+1$$

Simplify. The value x=\frac{15}{2} satisfies the equation.

$$\frac{17}{2}=\frac{17}{2}$$

Equation \sqrt{x^{2}+16}=x+1 has a unique solution.

$$x=\frac{15}{2}$$