Solve sqrt{2x+6}+sqrt{x+4}=1

$$\sqrt { 2 x + 6 } + \sqrt { x + 4 } = 1$$

Solve for x

$$x=-3$$

Solution Steps

Subtract \sqrt{x+4} from both sides of the equation.

$$\sqrt{2x+6}=1-\sqrt{x+4}$$

Square both sides of the equation.

$$\left(\sqrt{2x+6}\right)^{2}=\left(1-\sqrt{x+4}\right)^{2}$$

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

$$2x+6=\left(1-\sqrt{x+4}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{x+4}\right)^{2}.

$$2x+6=1-2\sqrt{x+4}+\left(\sqrt{x+4}\right)^{2}$$

Calculate \sqrt{x+4} to the power of 2 and get x+4.

$$2x+6=1-2\sqrt{x+4}+x+4$$

Add 1 and 4 to get 5.

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

Subtract 5+x from both sides of the equation.

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

To find the opposite of 5+x, find the opposite of each term.

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

Subtract 5 from 6 to get 1.

$$2x+1-x=-2\sqrt{x+4}$$

Combine 2x and -x to get x.

$$x+1=-2\sqrt{x+4}$$

Square both sides of the equation.

$$\left(x+1\right)^{2}=\left(-2\sqrt{x+4}\right)^{2}$$

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

$$x^{2}+2x+1=\left(-2\sqrt{x+4}\right)^{2}$$

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

$$x^{2}+2x+1=\left(-2\right)^{2}\left(\sqrt{x+4}\right)^{2}$$

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

$$x^{2}+2x+1=4\left(\sqrt{x+4}\right)^{2}$$

Calculate \sqrt{x+4} to the power of 2 and get x+4.

$$x^{2}+2x+1=4\left(x+4\right)$$

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

$$x^{2}+2x+1=4x+16$$

Subtract 4x from both sides.

$$x^{2}+2x+1-4x=16$$

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

$$x^{2}-2x+1=16$$

Subtract 16 from both sides.

$$x^{2}-2x+1-16=0$$

Subtract 16 from 1 to get -15.

$$x^{2}-2x-15=0$$

To solve the equation, factor x^{2}-2x-15 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-5\right)\left(x+3\right)$$

To find equation solutions, solve x-5=0 and x+3=0.

$$x=5 x=-3$$

Substitute 5 for x in the equation \sqrt{2x+6}+\sqrt{x+4}=1.

$$\sqrt{2\times 5+6}+\sqrt{5+4}=1$$

Simplify. The value x=5 does not satisfy the equation.

$$7=1$$

Substitute -3 for x in the equation \sqrt{2x+6}+\sqrt{x+4}=1.

$$\sqrt{2\left(-3\right)+6}+\sqrt{-3+4}=1$$

Simplify. The value x=-3 satisfies the equation.

$$1=1$$

Equation \sqrt{2x+6}=-\sqrt{x+4}+1 has a unique solution.

$$x=-3$$