Solve sqrt{x+2}=x-4

$$\sqrt{ x+2 } = x-4$$

Solve for x

$$x=7$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+2}\right)^{2}=\left(x-4\right)^{2}$$

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x+2=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$x+2-x^{2}=-8x+16$$

Add 8x to both sides.

$$x+2-x^{2}+8x=16$$

Combine x and 8x to get 9x.

$$9x+2-x^{2}=16$$

Subtract 16 from both sides.

$$9x+2-x^{2}-16=0$$

Subtract 16 from 2 to get -14.

$$9x-14-x^{2}=0$$

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

$$-x^{2}+9x-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-14\right)=14$$

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 14.

$$1,14 2,7$$

Calculate the sum for each pair.

$$1+14=15 2+7=9$$

The solution is the pair that gives sum 9.

$$a=7 b=2$$

Rewrite -x^{2}+9x-14 as \left(-x^{2}+7x\right)+\left(2x-14\right).

$$\left(-x^{2}+7x\right)+\left(2x-14\right)$$

Factor out -x in the first and 2 in the second group.

$$-x\left(x-7\right)+2\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(-x+2\right)$$

To find equation solutions, solve x-7=0 and -x+2=0.

$$x=7 x=2$$

Substitute 7 for x in the equation \sqrt{x+2}=x-4.

$$\sqrt{7+2}=7-4$$

Simplify. The value x=7 satisfies the equation.

$$3=3$$

Substitute 2 for x in the equation \sqrt{x+2}=x-4.

$$\sqrt{2+2}=2-4$$

Simplify. The value x=2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Equation \sqrt{x+2}=x-4 has a unique solution.

$$x=7$$