Solve sqrt{2x+8}-2sqrt{x+5}=-2

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

Solve for x

$$x=4$$

$$x=-4$$

Solution Steps

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

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

Square both sides of the equation.

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

Calculate \sqrt{2x+8} to the power of 2 and get 2x+8.

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

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

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

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

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

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

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

Add 4 and 20 to get 24.

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

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

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

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

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

Subtract 24 from 8 to get -16.

$$2x-16-4x=-8\sqrt{x+5}$$

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

$$-2x-16=-8\sqrt{x+5}$$

Square both sides of the equation.

$$\left(-2x-16\right)^{2}=\left(-8\sqrt{x+5}\right)^{2}$$

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

$$4x^{2}+64x+256=\left(-8\sqrt{x+5}\right)^{2}$$

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

$$4x^{2}+64x+256=\left(-8\right)^{2}\left(\sqrt{x+5}\right)^{2}$$

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

$$4x^{2}+64x+256=64\left(\sqrt{x+5}\right)^{2}$$

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

$$4x^{2}+64x+256=64\left(x+5\right)$$

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

$$4x^{2}+64x+256=64x+320$$

Subtract 64x from both sides.

$$4x^{2}+64x+256-64x=320$$

Combine 64x and -64x to get 0.

$$4x^{2}+256=320$$

Subtract 320 from both sides.

$$4x^{2}+256-320=0$$

Subtract 320 from 256 to get -64.

$$4x^{2}-64=0$$

Divide both sides by 4.

$$x^{2}-16=0$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

To find equation solutions, solve x-4=0 and x+4=0.

$$x=4 x=-4$$

Substitute 4 for x in the equation \sqrt{2x+8}-2\sqrt{x+5}=-2.

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

Simplify. The value x=4 satisfies the equation.

$$-2=-2$$

Substitute -4 for x in the equation \sqrt{2x+8}-2\sqrt{x+5}=-2.

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

Simplify. The value x=-4 satisfies the equation.

$$-2=-2$$

List all solutions of \sqrt{2x+8}=2\sqrt{x+5}-2.

$$x=4 x=-4$$