Solve sqrt{x+3}+sqrt{x+6}=sqrt{x+11}

$$\sqrt{ x+3 } + \sqrt{ x+6 } = \sqrt{ x+11 }$$

Solve for x

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+3}+\sqrt{x+6}\right)^{2}=\left(\sqrt{x+11}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+3}+\sqrt{x+6}\right)^{2}.

$$\left(\sqrt{x+3}\right)^{2}+2\sqrt{x+3}\sqrt{x+6}+\left(\sqrt{x+6}\right)^{2}=\left(\sqrt{x+11}\right)^{2}$$

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

$$x+3+2\sqrt{x+3}\sqrt{x+6}+\left(\sqrt{x+6}\right)^{2}=\left(\sqrt{x+11}\right)^{2}$$

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

$$x+3+2\sqrt{x+3}\sqrt{x+6}+x+6=\left(\sqrt{x+11}\right)^{2}$$

Combine x and x to get 2x.

$$2x+3+2\sqrt{x+3}\sqrt{x+6}+6=\left(\sqrt{x+11}\right)^{2}$$

Add 3 and 6 to get 9.

$$2x+9+2\sqrt{x+3}\sqrt{x+6}=\left(\sqrt{x+11}\right)^{2}$$

Calculate \sqrt{x+11} to the power of 2 and get x+11.

$$2x+9+2\sqrt{x+3}\sqrt{x+6}=x+11$$

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

$$2\sqrt{x+3}\sqrt{x+6}=x+11-\left(2x+9\right)$$

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

$$2\sqrt{x+3}\sqrt{x+6}=x+11-2x-9$$

Combine x and -2x to get -x.

$$2\sqrt{x+3}\sqrt{x+6}=-x+11-9$$

Subtract 9 from 11 to get 2.

$$2\sqrt{x+3}\sqrt{x+6}=-x+2$$

Square both sides of the equation.

$$\left(2\sqrt{x+3}\sqrt{x+6}\right)^{2}=\left(-x+2\right)^{2}$$

Expand \left(2\sqrt{x+3}\sqrt{x+6}\right)^{2}.

$$2^{2}\left(\sqrt{x+3}\right)^{2}\left(\sqrt{x+6}\right)^{2}=\left(-x+2\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

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

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

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

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

$$4\left(x+3\right)\left(x+6\right)=\left(-x+2\right)^{2}$$

Use the distributive property to multiply 4 by x+3.

$$\left(4x+12\right)\left(x+6\right)=\left(-x+2\right)^{2}$$

Apply the distributive property by multiplying each term of 4x+12 by each term of x+6.

$$4x^{2}+24x+12x+72=\left(-x+2\right)^{2}$$

Combine 24x and 12x to get 36x.

$$4x^{2}+36x+72=\left(-x+2\right)^{2}$$

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

$$4x^{2}+36x+72=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$4x^{2}+36x+72-x^{2}=-4x+4$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+36x+72=-4x+4$$

Add 4x to both sides.

$$3x^{2}+36x+72+4x=4$$

Combine 36x and 4x to get 40x.

$$3x^{2}+40x+72=4$$

Subtract 4 from both sides.

$$3x^{2}+40x+72-4=0$$

Subtract 4 from 72 to get 68.

$$3x^{2}+40x+68=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+68. To find a and b, set up a system to be solved.

$$a+b=40 ab=3\times 68=204$$

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 204.

$$1,204 2,102 3,68 4,51 6,34 12,17$$

Calculate the sum for each pair.

$$1+204=205 2+102=104 3+68=71 4+51=55 6+34=40 12+17=29$$

The solution is the pair that gives sum 40.

$$a=6 b=34$$

Rewrite 3x^{2}+40x+68 as \left(3x^{2}+6x\right)+\left(34x+68\right).

$$\left(3x^{2}+6x\right)+\left(34x+68\right)$$

Factor out 3x in the first and 34 in the second group.

$$3x\left(x+2\right)+34\left(x+2\right)$$

Factor out common term x+2 by using distributive property.

$$\left(x+2\right)\left(3x+34\right)$$

To find equation solutions, solve x+2=0 and 3x+34=0.

$$x=-2 x=-\frac{34}{3}$$

Substitute -\frac{34}{3} for x in the equation \sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11}. The expression \sqrt{-\frac{34}{3}+3} is undefined because the radicand cannot be negative.

$$\sqrt{-\frac{34}{3}+3}+\sqrt{-\frac{34}{3}+6}=\sqrt{-\frac{34}{3}+11}$$

Substitute -2 for x in the equation \sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11}.

$$\sqrt{-2+3}+\sqrt{-2+6}=\sqrt{-2+11}$$

Simplify. The value x=-2 satisfies the equation.

$$3=3$$

Equation \sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11} has a unique solution.

$$x=-2$$