Solve sqrt{x+24}=x-6

$$\sqrt{ x+24 } =x-6$$

Solve for x

$$x=12$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+24}\right)^{2}=\left(x-6\right)^{2}$$

Calculate \sqrt{x+24} to the power of 2 and get x+24.

$$x+24=\left(x-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x+24=x^{2}-12x+36$$

Subtract x^{2} from both sides.

$$x+24-x^{2}=-12x+36$$

Add 12x to both sides.

$$x+24-x^{2}+12x=36$$

Combine x and 12x to get 13x.

$$13x+24-x^{2}=36$$

Subtract 36 from both sides.

$$13x+24-x^{2}-36=0$$

Subtract 36 from 24 to get -12.

$$13x-12-x^{2}=0$$

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

$$-x^{2}+13x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-12\right)=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=12 b=1$$

Rewrite -x^{2}+13x-12 as \left(-x^{2}+12x\right)+\left(x-12\right).

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

Factor out -x in -x^{2}+12x.

$$-x\left(x-12\right)+x-12$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x+1\right)$$

To find equation solutions, solve x-12=0 and -x+1=0.

$$x=12 x=1$$

Substitute 12 for x in the equation \sqrt{x+24}=x-6.

$$\sqrt{12+24}=12-6$$

Simplify. The value x=12 satisfies the equation.

$$6=6$$

Substitute 1 for x in the equation \sqrt{x+24}=x-6.

$$\sqrt{1+24}=1-6$$

Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$5=-5$$

Equation \sqrt{x+24}=x-6 has a unique solution.

$$x=12$$