Solve sqrt{x+2}+x=10

$$\sqrt{ x+2 } +x=10$$

Solve for x

$$x=7$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{x+2}=10-x$$

Square both sides of the equation.

$$\left(\sqrt{x+2}\right)^{2}=\left(10-x\right)^{2}$$

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-x\right)^{2}.

$$x+2=100-20x+x^{2}$$

Subtract 100 from both sides.

$$x+2-100=-20x+x^{2}$$

Subtract 100 from 2 to get -98.

$$x-98=-20x+x^{2}$$

Add 20x to both sides.

$$x-98+20x=x^{2}$$

Combine x and 20x to get 21x.

$$21x-98=x^{2}$$

Subtract x^{2} from both sides.

$$21x-98-x^{2}=0$$

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

$$-x^{2}+21x-98=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-98. To find a and b, set up a system to be solved.

$$a+b=21 ab=-\left(-98\right)=98$$

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 98.

$$1,98 2,49 7,14$$

Calculate the sum for each pair.

$$1+98=99 2+49=51 7+14=21$$

The solution is the pair that gives sum 21.

$$a=14 b=7$$

Rewrite -x^{2}+21x-98 as \left(-x^{2}+14x\right)+\left(7x-98\right).

$$\left(-x^{2}+14x\right)+\left(7x-98\right)$$

Factor out -x in the first and 7 in the second group.

$$-x\left(x-14\right)+7\left(x-14\right)$$

Factor out common term x-14 by using distributive property.

$$\left(x-14\right)\left(-x+7\right)$$

To find equation solutions, solve x-14=0 and -x+7=0.

$$x=14 x=7$$

Substitute 14 for x in the equation \sqrt{x+2}+x=10.

$$\sqrt{14+2}+14=10$$

Simplify. The value x=14 does not satisfy the equation.

$$18=10$$

Substitute 7 for x in the equation \sqrt{x+2}+x=10.

$$\sqrt{7+2}+7=10$$

Simplify. The value x=7 satisfies the equation.

$$10=10$$

Equation \sqrt{x+2}=10-x has a unique solution.

$$x=7$$