Solve sqrt{2x+6}-2sqrt{x+5}=-2

$$\sqrt { 2 x + 6 } - 2 \sqrt { x + 5 } = - 2$$

Solve for x

$$x=-1$$

Solution Steps

Subtract -2\sqrt{x+5} from both sides of the equation.

$$\sqrt{2x+6}=-2+2\sqrt{x+5}$$

Square both sides of the equation.

$$\left(\sqrt{2x+6}\right)^{2}=\left(-2+2\sqrt{x+5}\right)^{2}$$

Calculate \sqrt{2x+6} to the power of 2 and get 2x+6.

$$2x+6=\left(-2+2\sqrt{x+5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2+2\sqrt{x+5}\right)^{2}.

$$2x+6=4-8\sqrt{x+5}+4\left(\sqrt{x+5}\right)^{2}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$2x+6=4-8\sqrt{x+5}+4\left(x+5\right)$$

Use the distributive property to multiply 4 by x+5.

$$2x+6=4-8\sqrt{x+5}+4x+20$$

Add 4 and 20 to get 24.

$$2x+6=24-8\sqrt{x+5}+4x$$

Subtract 24+4x from both sides of the equation.

$$2x+6-\left(24+4x\right)=-8\sqrt{x+5}$$

To find the opposite of 24+4x, find the opposite of each term.

$$2x+6-24-4x=-8\sqrt{x+5}$$

Subtract 24 from 6 to get -18.

$$2x-18-4x=-8\sqrt{x+5}$$

Combine 2x and -4x to get -2x.

$$-2x-18=-8\sqrt{x+5}$$

Square both sides of the equation.

$$\left(-2x-18\right)^{2}=\left(-8\sqrt{x+5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2x-18\right)^{2}.

$$4x^{2}+72x+324=\left(-8\sqrt{x+5}\right)^{2}$$

Expand \left(-8\sqrt{x+5}\right)^{2}.

$$4x^{2}+72x+324=\left(-8\right)^{2}\left(\sqrt{x+5}\right)^{2}$$

Calculate -8 to the power of 2 and get 64.

$$4x^{2}+72x+324=64\left(\sqrt{x+5}\right)^{2}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$4x^{2}+72x+324=64\left(x+5\right)$$

Use the distributive property to multiply 64 by x+5.

$$4x^{2}+72x+324=64x+320$$

Subtract 64x from both sides.

$$4x^{2}+72x+324-64x=320$$

Combine 72x and -64x to get 8x.

$$4x^{2}+8x+324=320$$

Subtract 320 from both sides.

$$4x^{2}+8x+324-320=0$$

Subtract 320 from 324 to get 4.

$$4x^{2}+8x+4=0$$

Divide both sides by 4.

$$x^{2}+2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\times 1=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite x^{2}+2x+1 as \left(x^{2}+x\right)+\left(x+1\right).

$$\left(x^{2}+x\right)+\left(x+1\right)$$

Factor out x in x^{2}+x.

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

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

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

Rewrite as a binomial square.

$$\left(x+1\right)^{2}$$

To find equation solution, solve x+1=0.

$$x=-1$$

Substitute -1 for x in the equation \sqrt{2x+6}-2\sqrt{x+5}=-2.

$$\sqrt{2\left(-1\right)+6}-2\sqrt{-1+5}=-2$$

Simplify. The value x=-1 satisfies the equation.

$$-2=-2$$

Equation \sqrt{2x+6}=2\sqrt{x+5}-2 has a unique solution.

$$x=-1$$