Solve sqrt{16+x}=x-4

$$\sqrt { 16 + x } = x - 4$$

Solve for x

$$x=9$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{16+x}\right)^{2}=\left(x-4\right)^{2}$$

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

$$16+x=\left(x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

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

Subtract x^{2} from both sides.

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

Add 8x to both sides.

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

Combine x and 8x to get 9x.

$$16+9x-x^{2}=16$$

Subtract 16 from both sides.

$$16+9x-x^{2}-16=0$$

Subtract 16 from 16 to get 0.

$$9x-x^{2}=0$$

Factor out x.

$$x\left(9-x\right)=0$$

To find equation solutions, solve x=0 and 9-x=0.

$$x=0 x=9$$

Substitute 0 for x in the equation \sqrt{16+x}=x-4.

$$\sqrt{16+0}=0-4$$

Simplify. The value x=0 does not satisfy the equation because the left and the right hand side have opposite signs.

$$4=-4$$

Substitute 9 for x in the equation \sqrt{16+x}=x-4.

$$\sqrt{16+9}=9-4$$

Simplify. The value x=9 satisfies the equation.

$$5=5$$

Equation \sqrt{x+16}=x-4 has a unique solution.

$$x=9$$