Solve sqrt{3-x}-sqrt{16+2x}=sqrt{x+7}

$$\sqrt { 3 - x } - \sqrt { 16 + 2 x } = \sqrt { x + 7 }$$

Solve for x

$$x=-6$$

Solution Steps

Subtract -\sqrt{16+2x} from both sides of the equation.

$$\sqrt{3-x}=\sqrt{x+7}+\sqrt{16+2x}$$

Square both sides of the equation.

$$\left(\sqrt{3-x}\right)^{2}=\left(\sqrt{x+7}+\sqrt{16+2x}\right)^{2}$$

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

$$3-x=\left(\sqrt{x+7}+\sqrt{16+2x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+7}+\sqrt{16+2x}\right)^{2}.

$$3-x=\left(\sqrt{x+7}\right)^{2}+2\sqrt{x+7}\sqrt{16+2x}+\left(\sqrt{16+2x}\right)^{2}$$

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

$$3-x=x+7+2\sqrt{x+7}\sqrt{16+2x}+\left(\sqrt{16+2x}\right)^{2}$$

Calculate \sqrt{16+2x} to the power of 2 and get 16+2x.

$$3-x=x+7+2\sqrt{x+7}\sqrt{16+2x}+16+2x$$

Add 7 and 16 to get 23.

$$3-x=x+23+2\sqrt{x+7}\sqrt{16+2x}+2x$$

Combine x and 2x to get 3x.

$$3-x=3x+23+2\sqrt{x+7}\sqrt{16+2x}$$

Subtract 3x+23 from both sides of the equation.

$$3-x-\left(3x+23\right)=2\sqrt{x+7}\sqrt{16+2x}$$

To find the opposite of 3x+23, find the opposite of each term.

$$3-x-3x-23=2\sqrt{x+7}\sqrt{16+2x}$$

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

$$3-4x-23=2\sqrt{x+7}\sqrt{16+2x}$$

Subtract 23 from 3 to get -20.

$$-20-4x=2\sqrt{x+7}\sqrt{16+2x}$$

Square both sides of the equation.

$$\left(-20-4x\right)^{2}=\left(2\sqrt{x+7}\sqrt{16+2x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-20-4x\right)^{2}.

$$400+160x+16x^{2}=\left(2\sqrt{x+7}\sqrt{16+2x}\right)^{2}$$

Expand \left(2\sqrt{x+7}\sqrt{16+2x}\right)^{2}.

$$400+160x+16x^{2}=2^{2}\left(\sqrt{x+7}\right)^{2}\left(\sqrt{16+2x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$400+160x+16x^{2}=4\left(\sqrt{x+7}\right)^{2}\left(\sqrt{16+2x}\right)^{2}$$

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

$$400+160x+16x^{2}=4\left(x+7\right)\left(\sqrt{16+2x}\right)^{2}$$

Calculate \sqrt{16+2x} to the power of 2 and get 16+2x.

$$400+160x+16x^{2}=4\left(x+7\right)\left(16+2x\right)$$

Use the distributive property to multiply 4 by x+7.

$$400+160x+16x^{2}=\left(4x+28\right)\left(16+2x\right)$$

Apply the distributive property by multiplying each term of 4x+28 by each term of 16+2x.

$$400+160x+16x^{2}=64x+8x^{2}+448+56x$$

Combine 64x and 56x to get 120x.

$$400+160x+16x^{2}=120x+8x^{2}+448$$

Subtract 120x from both sides.

$$400+160x+16x^{2}-120x=8x^{2}+448$$

Combine 160x and -120x to get 40x.

$$400+40x+16x^{2}=8x^{2}+448$$

Subtract 8x^{2} from both sides.

$$400+40x+16x^{2}-8x^{2}=448$$

Combine 16x^{2} and -8x^{2} to get 8x^{2}.

$$400+40x+8x^{2}=448$$

Subtract 448 from both sides.

$$400+40x+8x^{2}-448=0$$

Subtract 448 from 400 to get -48.

$$-48+40x+8x^{2}=0$$

Divide both sides by 8.

$$-6+5x+x^{2}=0$$

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

$$x^{2}+5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite x^{2}+5x-6 as \left(x^{2}-x\right)+\left(6x-6\right).

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

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

$$x\left(x-1\right)+6\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+6\right)$$

To find equation solutions, solve x-1=0 and x+6=0.

$$x=1 x=-6$$

Substitute 1 for x in the equation \sqrt{3-x}-\sqrt{16+2x}=\sqrt{x+7}.

$$\sqrt{3-1}-\sqrt{16+2\times 1}=\sqrt{1+7}$$

Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2\times 2^{\frac{1}{2}}=2\times 2^{\frac{1}{2}}$$

Substitute -6 for x in the equation \sqrt{3-x}-\sqrt{16+2x}=\sqrt{x+7}.

$$\sqrt{3-\left(-6\right)}-\sqrt{16+2\left(-6\right)}=\sqrt{-6+7}$$

Simplify. The value x=-6 satisfies the equation.

$$1=1$$

Equation \sqrt{3-x}=\sqrt{x+7}+\sqrt{2x+16} has a unique solution.

$$x=-6$$