Solve sqrt{x}=x-12

$$\sqrt{ x } = x-12$$

Solve for x

$$x=16$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x}\right)^{2}=\left(x-12\right)^{2}$$

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

$$x=\left(x-12\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-12\right)^{2}.

$$x=x^{2}-24x+144$$

Subtract x^{2} from both sides.

$$x-x^{2}=-24x+144$$

Add 24x to both sides.

$$x-x^{2}+24x=144$$

Combine x and 24x to get 25x.

$$25x-x^{2}=144$$

Subtract 144 from both sides.

$$25x-x^{2}-144=0$$

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

$$-x^{2}+25x-144=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-144. To find a and b, set up a system to be solved.

$$a+b=25 ab=-\left(-144\right)=144$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 144.

$$1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12$$

Calculate the sum for each pair.

$$1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24$$

The solution is the pair that gives sum 25.

$$a=16 b=9$$

Rewrite -x^{2}+25x-144 as \left(-x^{2}+16x\right)+\left(9x-144\right).

$$\left(-x^{2}+16x\right)+\left(9x-144\right)$$

Factor out -x in the first and 9 in the second group.

$$-x\left(x-16\right)+9\left(x-16\right)$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(-x+9\right)$$

To find equation solutions, solve x-16=0 and -x+9=0.

$$x=16 x=9$$

Substitute 16 for x in the equation \sqrt{x}=x-12.

$$\sqrt{16}=16-12$$

Simplify. The value x=16 satisfies the equation.

$$4=4$$

Substitute 9 for x in the equation \sqrt{x}=x-12.

$$\sqrt{9}=9-12$$

Simplify. The value x=9 does not satisfy the equation because the left and the right hand side have opposite signs.

$$3=-3$$

Equation \sqrt{x}=x-12 has a unique solution.

$$x=16$$