Solve sqrt{x+6}+6=x

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

Solve for x

$$x=10$$

Solution Steps

Subtract 6 from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{x+6} to the power of 2 and get x+6.

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

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

$$x+6=x^{2}-12x+36$$

Subtract x^{2} from both sides.

$$x+6-x^{2}=-12x+36$$

Add 12x to both sides.

$$x+6-x^{2}+12x=36$$

Combine x and 12x to get 13x.

$$13x+6-x^{2}=36$$

Subtract 36 from both sides.

$$13x+6-x^{2}-36=0$$

Subtract 36 from 6 to get -30.

$$13x-30-x^{2}=0$$

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

$$-x^{2}+13x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-30\right)=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=10 b=3$$

Rewrite -x^{2}+13x-30 as \left(-x^{2}+10x\right)+\left(3x-30\right).

$$\left(-x^{2}+10x\right)+\left(3x-30\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-10\right)+3\left(x-10\right)$$

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

$$\left(x-10\right)\left(-x+3\right)$$

To find equation solutions, solve x-10=0 and -x+3=0.

$$x=10 x=3$$

Substitute 10 for x in the equation \sqrt{x+6}+6=x.

$$\sqrt{10+6}+6=10$$

Simplify. The value x=10 satisfies the equation.

$$10=10$$

Substitute 3 for x in the equation \sqrt{x+6}+6=x.

$$\sqrt{3+6}+6=3$$

Simplify. The value x=3 does not satisfy the equation.

$$9=3$$

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

$$x=10$$