Solve sqrt{x+7}-sqrt{3-x}=2

$$\sqrt{ x+7 } - \sqrt{ 3-x } =2$$

Solve for x

$$x=2$$

Solution Steps

Subtract -\sqrt{3-x} from both sides of the equation.

$$\sqrt{x+7}=2+\sqrt{3-x}$$

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{3-x}\right)^{2}.

$$x+7=4+4\sqrt{3-x}+\left(\sqrt{3-x}\right)^{2}$$

Calculate \sqrt{3-x} to the power of 2 and get 3-x.

$$x+7=4+4\sqrt{3-x}+3-x$$

Add 4 and 3 to get 7.

$$x+7=7+4\sqrt{3-x}-x$$

Subtract 7-x from both sides of the equation.

$$x+7-\left(7-x\right)=4\sqrt{3-x}$$

To find the opposite of 7-x, find the opposite of each term.

$$x+7-7+x=4\sqrt{3-x}$$

Subtract 7 from 7 to get 0.

$$x+x=4\sqrt{3-x}$$

Combine x and x to get 2x.

$$2x=4\sqrt{3-x}$$

Square both sides of the equation.

$$\left(2x\right)^{2}=\left(4\sqrt{3-x}\right)^{2}$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}=\left(4\sqrt{3-x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=\left(4\sqrt{3-x}\right)^{2}$$

Expand \left(4\sqrt{3-x}\right)^{2}.

$$4x^{2}=4^{2}\left(\sqrt{3-x}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$4x^{2}=16\left(\sqrt{3-x}\right)^{2}$$

Calculate \sqrt{3-x} to the power of 2 and get 3-x.

$$4x^{2}=16\left(3-x\right)$$

Use the distributive property to multiply 16 by 3-x.

$$4x^{2}=48-16x$$

Subtract 48 from both sides.

$$4x^{2}-48=-16x$$

Add 16x to both sides.

$$4x^{2}-48+16x=0$$

Divide both sides by 4.

$$x^{2}-12+4x=0$$

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

$$x^{2}+4x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite x^{2}+4x-12 as \left(x^{2}-2x\right)+\left(6x-12\right).

$$\left(x^{2}-2x\right)+\left(6x-12\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-2\right)+6\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+6\right)$$

To find equation solutions, solve x-2=0 and x+6=0.

$$x=2 x=-6$$

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

$$\sqrt{2+7}-\sqrt{3-2}=2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Substitute -6 for x in the equation \sqrt{x+7}-\sqrt{3-x}=2.

$$\sqrt{-6+7}-\sqrt{3-\left(-6\right)}=2$$

Simplify. The value x=-6 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

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

$$\sqrt{2+7}-\sqrt{3-2}=2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Equation \sqrt{x+7}=\sqrt{3-x}+2 has a unique solution.

$$x=2$$