Solve sqrt{x+7}-sqrt{13-x}=2

$$\sqrt { x + 7 } - \sqrt { 13 - x } = 2$$

Solve for x

$$x=9$$

Solution Steps

Subtract -\sqrt{13-x} from both sides of the equation.

$$\sqrt{x+7}=2+\sqrt{13-x}$$

Square both sides of the equation.

$$\left(\sqrt{x+7}\right)^{2}=\left(2+\sqrt{13-x}\right)^{2}$$

Calculate \sqrt{x+7} to the power of 2 and get x+7.

$$x+7=\left(2+\sqrt{13-x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{13-x}\right)^{2}.

$$x+7=4+4\sqrt{13-x}+\left(\sqrt{13-x}\right)^{2}$$

Calculate \sqrt{13-x} to the power of 2 and get 13-x.

$$x+7=4+4\sqrt{13-x}+13-x$$

Add 4 and 13 to get 17.

$$x+7=17+4\sqrt{13-x}-x$$

Subtract 17-x from both sides of the equation.

$$x+7-\left(17-x\right)=4\sqrt{13-x}$$

To find the opposite of 17-x, find the opposite of each term.

$$x+7-17+x=4\sqrt{13-x}$$

Subtract 17 from 7 to get -10.

$$x-10+x=4\sqrt{13-x}$$

Combine x and x to get 2x.

$$2x-10=4\sqrt{13-x}$$

Square both sides of the equation.

$$\left(2x-10\right)^{2}=\left(4\sqrt{13-x}\right)^{2}$$

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

$$4x^{2}-40x+100=\left(4\sqrt{13-x}\right)^{2}$$

Expand \left(4\sqrt{13-x}\right)^{2}.

$$4x^{2}-40x+100=4^{2}\left(\sqrt{13-x}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$4x^{2}-40x+100=16\left(\sqrt{13-x}\right)^{2}$$

Calculate \sqrt{13-x} to the power of 2 and get 13-x.

$$4x^{2}-40x+100=16\left(13-x\right)$$

Use the distributive property to multiply 16 by 13-x.

$$4x^{2}-40x+100=208-16x$$

Subtract 208 from both sides.

$$4x^{2}-40x+100-208=-16x$$

Subtract 208 from 100 to get -108.

$$4x^{2}-40x-108=-16x$$

Add 16x to both sides.

$$4x^{2}-40x-108+16x=0$$

Combine -40x and 16x to get -24x.

$$4x^{2}-24x-108=0$$

Divide both sides by 4.

$$x^{2}-6x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-27\right)=-27$$

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 -27.

$$1,-27 3,-9$$

Calculate the sum for each pair.

$$1-27=-26 3-9=-6$$

The solution is the pair that gives sum -6.

$$a=-9 b=3$$

Rewrite x^{2}-6x-27 as \left(x^{2}-9x\right)+\left(3x-27\right).

$$\left(x^{2}-9x\right)+\left(3x-27\right)$$

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

$$x\left(x-9\right)+3\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x+3\right)$$

To find equation solutions, solve x-9=0 and x+3=0.

$$x=9 x=-3$$

Substitute 9 for x in the equation \sqrt{x+7}-\sqrt{13-x}=2.

$$\sqrt{9+7}-\sqrt{13-9}=2$$

Simplify. The value x=9 satisfies the equation.

$$2=2$$

Substitute -3 for x in the equation \sqrt{x+7}-\sqrt{13-x}=2.

$$\sqrt{-3+7}-\sqrt{13-\left(-3\right)}=2$$

Simplify. The value x=-3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Substitute 9 for x in the equation \sqrt{x+7}-\sqrt{13-x}=2.

$$\sqrt{9+7}-\sqrt{13-9}=2$$

Simplify. The value x=9 satisfies the equation.

$$2=2$$

Equation \sqrt{x+7}=\sqrt{13-x}+2 has a unique solution.

$$x=9$$