Solve sqrt{x+14}-x=2

$$\sqrt { x + 14 } - x = 2$$

Solve for x

$$x=2$$

Solution Steps

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

$$\left(\sqrt{x+14}\right)^{2}=\left(2+x\right)^{2}$$

Calculate \sqrt{x+14} to the power of 2 and get x+14.

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

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

$$x+14=4+4x+x^{2}$$

Subtract 4 from both sides.

$$x+14-4=4x+x^{2}$$

Subtract 4 from 14 to get 10.

$$x+10=4x+x^{2}$$

Subtract 4x from both sides.

$$x+10-4x=x^{2}$$

Combine x and -4x to get -3x.

$$-3x+10=x^{2}$$

Subtract x^{2} from both sides.

$$-3x+10-x^{2}=0$$

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

$$-x^{2}-3x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-10=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=2 b=-5$$

Rewrite -x^{2}-3x+10 as \left(-x^{2}+2x\right)+\left(-5x+10\right).

$$\left(-x^{2}+2x\right)+\left(-5x+10\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(-x+2\right)+5\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(x+5\right)$$

To find equation solutions, solve -x+2=0 and x+5=0.

$$x=2 x=-5$$

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

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

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Substitute -5 for x in the equation \sqrt{x+14}-x=2.

$$\sqrt{-5+14}-\left(-5\right)=2$$

Simplify. The value x=-5 does not satisfy the equation.

$$8=2$$

Equation \sqrt{x+14}=x+2 has a unique solution.

$$x=2$$