Solve sqrt{20+x}=x

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

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

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

Subtract x^{2} from both sides.

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

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

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

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.

$$a+b=1 ab=-20=-20$$

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 1.

$$a=5 b=-4$$

Rewrite -x^{2}+x+20 as \left(-x^{2}+5x\right)+\left(-4x+20\right).

$$\left(-x^{2}+5x\right)+\left(-4x+20\right)$$

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

$$-x\left(x-5\right)-4\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x-4\right)$$

To find equation solutions, solve x-5=0 and -x-4=0.

$$x=5 x=-4$$

Substitute 5 for x in the equation \sqrt{20+x}=x.

$$\sqrt{20+5}=5$$

Simplify. The value x=5 satisfies the equation.

$$5=5$$

Substitute -4 for x in the equation \sqrt{20+x}=x.

$$\sqrt{20-4}=-4$$

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{x+20}=x has a unique solution.

$$x=5$$