Solve sqrt{8x+48}=x

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

Solve for x

$$x=12$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{8x+48} to the power of 2 and get 8x+48.

$$8x+48=x^{2}$$

Subtract x^{2} from both sides.

$$8x+48-x^{2}=0$$

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

$$-x^{2}+8x+48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+48. To find a and b, set up a system to be solved.

$$a+b=8 ab=-48=-48$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=12 b=-4$$

Rewrite -x^{2}+8x+48 as \left(-x^{2}+12x\right)+\left(-4x+48\right).

$$\left(-x^{2}+12x\right)+\left(-4x+48\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(x-12\right)-4\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x-4\right)$$

To find equation solutions, solve x-12=0 and -x-4=0.

$$x=12 x=-4$$

Substitute 12 for x in the equation \sqrt{8x+48}=x.

$$\sqrt{8\times 12+48}=12$$

Simplify. The value x=12 satisfies the equation.

$$12=12$$

Substitute -4 for x in the equation \sqrt{8x+48}=x.

$$\sqrt{8\left(-4\right)+48}=-4$$

Simplify. The value x=-4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$4=-4$$

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

$$x=12$$