Solve sqrt{x+48}+8=x

$$\sqrt { x + 48 } + 8 = x$$

Solve for x

$$x=16$$

Solution Steps

Subtract 8 from both sides of the equation.

$$\sqrt{x+48}=x-8$$

Square both sides of the equation.

$$\left(\sqrt{x+48}\right)^{2}=\left(x-8\right)^{2}$$

Calculate \sqrt{x+48} to the power of 2 and get x+48.

$$x+48=\left(x-8\right)^{2}$$

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

$$x+48=x^{2}-16x+64$$

Subtract x^{2} from both sides.

$$x+48-x^{2}=-16x+64$$

Add 16x to both sides.

$$x+48-x^{2}+16x=64$$

Combine x and 16x to get 17x.

$$17x+48-x^{2}=64$$

Subtract 64 from both sides.

$$17x+48-x^{2}-64=0$$

Subtract 64 from 48 to get -16.

$$17x-16-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+17x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=17 ab=-\left(-16\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 17.

$$a=16 b=1$$

Rewrite -x^{2}+17x-16 as \left(-x^{2}+16x\right)+\left(x-16\right).

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

Factor out -x in -x^{2}+16x.

$$-x\left(x-16\right)+x-16$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(-x+1\right)$$

To find equation solutions, solve x-16=0 and -x+1=0.

$$x=16 x=1$$

Substitute 16 for x in the equation \sqrt{x+48}+8=x.

$$\sqrt{16+48}+8=16$$

Simplify. The value x=16 satisfies the equation.

$$16=16$$

Substitute 1 for x in the equation \sqrt{x+48}+8=x.

$$\sqrt{1+48}+8=1$$

Simplify. The value x=1 does not satisfy the equation.

$$15=1$$

Equation \sqrt{x+48}=x-8 has a unique solution.

$$x=16$$