Solve sqrt{20-x}=x

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

Solve for x

$$x=4$$

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=4 b=-5$$

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

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

Factor out x in the first and 5 in the second group.

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

Factor out common term -x+4 by using distributive property.

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

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

$$x=4 x=-5$$

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

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

Simplify. The value x=4 satisfies the equation.

$$4=4$$

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

$$\sqrt{20-\left(-5\right)}=-5$$

Simplify. The value x=-5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$5=-5$$

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

$$x=4$$