Solve sqrt{8x+20}=x

$$\sqrt { 8 x + 20 } = x$$

Solve for x

$$x=10$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{8x+20}\right)^{2}=x^{2}$$

Calculate \sqrt{8x+20} to the power of 2 and get 8x+20.

$$8x+20=x^{2}$$

Subtract x^{2} from both sides.

$$8x+20-x^{2}=0$$

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

$$-x^{2}+8x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=8 ab=-20=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 8.

$$a=10 b=-2$$

Rewrite -x^{2}+8x+20 as \left(-x^{2}+10x\right)+\left(-2x+20\right).

$$\left(-x^{2}+10x\right)+\left(-2x+20\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-10\right)-2\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(-x-2\right)$$

To find equation solutions, solve x-10=0 and -x-2=0.

$$x=10 x=-2$$

Substitute 10 for x in the equation \sqrt{8x+20}=x.

$$\sqrt{8\times 10+20}=10$$

Simplify. The value x=10 satisfies the equation.

$$10=10$$

Substitute -2 for x in the equation \sqrt{8x+20}=x.

$$\sqrt{8\left(-2\right)+20}=-2$$

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Equation \sqrt{8x+20}=x has a unique solution.

$$x=10$$