Solve sqrt{16+x^2}=8-x

$$\sqrt { 16 + x ^ { 2 } } = 8 - x$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{16+x^{2}}\right)^{2}=\left(8-x\right)^{2}$$

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

$$16+x^{2}=\left(8-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$16+x^{2}=64-16x+x^{2}$$

Add 16x to both sides.

$$16+x^{2}+16x=64+x^{2}$$

Subtract x^{2} from both sides.

$$16+x^{2}+16x-x^{2}=64$$

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

$$16+16x=64$$

Subtract 16 from both sides.

$$16x=64-16$$

Subtract 16 from 64 to get 48.

$$16x=48$$

Divide both sides by 16.

$$x=\frac{48}{16}$$

Divide 48 by 16 to get 3.

$$x=3$$

Substitute 3 for x in the equation \sqrt{16+x^{2}}=8-x.

$$\sqrt{16+3^{2}}=8-3$$

Simplify. The value x=3 satisfies the equation.

$$5=5$$

Equation \sqrt{x^{2}+16}=8-x has a unique solution.

$$x=3$$