Solve sqrt{3x+13}+sqrt{x+4}=1

$$\sqrt { 3 x + 13 } + \sqrt { x + 4 } = 1$$

Solve for x

$$x=-4$$

Solution Steps

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

$$\sqrt{3x+13}=1-\sqrt{x+4}$$

Square both sides of the equation.

$$\left(\sqrt{3x+13}\right)^{2}=\left(1-\sqrt{x+4}\right)^{2}$$

Calculate \sqrt{3x+13} to the power of 2 and get 3x+13.

$$3x+13=\left(1-\sqrt{x+4}\right)^{2}$$

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

$$3x+13=1-2\sqrt{x+4}+\left(\sqrt{x+4}\right)^{2}$$

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

$$3x+13=1-2\sqrt{x+4}+x+4$$

Add 1 and 4 to get 5.

$$3x+13=5-2\sqrt{x+4}+x$$

Subtract 5+x from both sides of the equation.

$$3x+13-\left(5+x\right)=-2\sqrt{x+4}$$

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

$$3x+13-5-x=-2\sqrt{x+4}$$

Subtract 5 from 13 to get 8.

$$3x+8-x=-2\sqrt{x+4}$$

Combine 3x and -x to get 2x.

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

Square both sides of the equation.

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

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

$$4x^{2}+32x+64=\left(-2\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=\left(-2\right)^{2}\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(\sqrt{x+4}\right)^{2}$$

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

$$4x^{2}+32x+64=4\left(x+4\right)$$

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

$$4x^{2}+32x+64=4x+16$$

Subtract 4x from both sides.

$$4x^{2}+32x+64-4x=16$$

Combine 32x and -4x to get 28x.

$$4x^{2}+28x+64=16$$

Subtract 16 from both sides.

$$4x^{2}+28x+64-16=0$$

Subtract 16 from 64 to get 48.

$$4x^{2}+28x+48=0$$

Divide both sides by 4.

$$x^{2}+7x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite x^{2}+7x+12 as \left(x^{2}+3x\right)+\left(4x+12\right).

$$\left(x^{2}+3x\right)+\left(4x+12\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x+3\right)+4\left(x+3\right)$$

Factor out common term x+3 by using distributive property.

$$\left(x+3\right)\left(x+4\right)$$

To find equation solutions, solve x+3=0 and x+4=0.

$$x=-3 x=-4$$

Substitute -3 for x in the equation \sqrt{3x+13}+\sqrt{x+4}=1.

$$\sqrt{3\left(-3\right)+13}+\sqrt{-3+4}=1$$

Simplify. The value x=-3 does not satisfy the equation.

$$3=1$$

Substitute -4 for x in the equation \sqrt{3x+13}+\sqrt{x+4}=1.

$$\sqrt{3\left(-4\right)+13}+\sqrt{-4+4}=1$$

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

$$1=1$$

Equation \sqrt{3x+13}=-\sqrt{x+4}+1 has a unique solution.

$$x=-4$$