Solve sqrt{2x+48}=x

$$\sqrt { 2 x + 48 } = x$$

Solve for x

$$x=8$$

Solution Steps

Square both sides of the equation.

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

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

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

Subtract x^{2} from both sides.

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

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

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

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.

$$a+b=2 ab=-48=-48$$

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 2.

$$a=8 b=-6$$

Rewrite -x^{2}+2x+48 as \left(-x^{2}+8x\right)+\left(-6x+48\right).

$$\left(-x^{2}+8x\right)+\left(-6x+48\right)$$

Factor out -x in the first and -6 in the second group.

$$-x\left(x-8\right)-6\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x-6\right)$$

To find equation solutions, solve x-8=0 and -x-6=0.

$$x=8 x=-6$$

Substitute 8 for x in the equation \sqrt{2x+48}=x.

$$\sqrt{2\times 8+48}=8$$

Simplify. The value x=8 satisfies the equation.

$$8=8$$

Substitute -6 for x in the equation \sqrt{2x+48}=x.

$$\sqrt{2\left(-6\right)+48}=-6$$

Simplify. The value x=-6 does not satisfy the equation because the left and the right hand side have opposite signs.

$$6=-6$$

Equation \sqrt{2x+48}=x has a unique solution.

$$x=8$$