Solve sqrt{x+5}+sqrt{2x+8}=1

$$\sqrt { x + 5 } + \sqrt { 2 x + 8 } = 1$$

Solve for x

$$x=-4$$

Solution Steps

Subtract \sqrt{2x+8} from both sides of the equation.

$$\sqrt{x+5}=1-\sqrt{2x+8}$$

Square both sides of the equation.

$$\left(\sqrt{x+5}\right)^{2}=\left(1-\sqrt{2x+8}\right)^{2}$$

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

$$x+5=\left(1-\sqrt{2x+8}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{2x+8}\right)^{2}.

$$x+5=1-2\sqrt{2x+8}+\left(\sqrt{2x+8}\right)^{2}$$

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

$$x+5=1-2\sqrt{2x+8}+2x+8$$

Add 1 and 8 to get 9.

$$x+5=9-2\sqrt{2x+8}+2x$$

Subtract 9+2x from both sides of the equation.

$$x+5-\left(9+2x\right)=-2\sqrt{2x+8}$$

To find the opposite of 9+2x, find the opposite of each term.

$$x+5-9-2x=-2\sqrt{2x+8}$$

Subtract 9 from 5 to get -4.

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

Combine x and -2x to get -x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate -2 to the power of 2 and get 4.

$$x^{2}+8x+16=4\left(\sqrt{2x+8}\right)^{2}$$

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

$$x^{2}+8x+16=4\left(2x+8\right)$$

Use the distributive property to multiply 4 by 2x+8.

$$x^{2}+8x+16=8x+32$$

Subtract 8x from both sides.

$$x^{2}+8x+16-8x=32$$

Combine 8x and -8x to get 0.

$$x^{2}+16=32$$

Subtract 32 from both sides.

$$x^{2}+16-32=0$$

Subtract 32 from 16 to get -16.

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

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{x+5}+\sqrt{2x+8}=1.

$$\sqrt{4+5}+\sqrt{2\times 4+8}=1$$

Simplify. The value x=4 does not satisfy the equation.

$$7=1$$

Substitute -4 for x in the equation \sqrt{x+5}+\sqrt{2x+8}=1.

$$\sqrt{-4+5}+\sqrt{2\left(-4\right)+8}=1$$

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

$$1=1$$

Equation \sqrt{x+5}=-\sqrt{2x+8}+1 has a unique solution.

$$x=-4$$