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

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

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

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

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

Calculate \sqrt{x+1} to the power of 2 and get x+1.

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

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

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

Combine x and x to get 2x.

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

Subtract 2 from 1 to get -1.

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

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

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

Subtract 2x-1 from both sides of the equation.

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

To find the opposite of 2x-1, find the opposite of each term.

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

Combine x and -2x to get -x.

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

Add 6 and 1 to get 7.

$$2\sqrt{x+1}\sqrt{x-2}=-x+7$$

Square both sides of the equation.

$$\left(2\sqrt{x+1}\sqrt{x-2}\right)^{2}=\left(-x+7\right)^{2}$$

Expand \left(2\sqrt{x+1}\sqrt{x-2}\right)^{2}.

$$2^{2}\left(\sqrt{x+1}\right)^{2}\left(\sqrt{x-2}\right)^{2}=\left(-x+7\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x+1}\right)^{2}\left(\sqrt{x-2}\right)^{2}=\left(-x+7\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$4\left(x+1\right)\left(\sqrt{x-2}\right)^{2}=\left(-x+7\right)^{2}$$

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

$$4\left(x+1\right)\left(x-2\right)=\left(-x+7\right)^{2}$$

Use the distributive property to multiply 4 by x+1.

$$\left(4x+4\right)\left(x-2\right)=\left(-x+7\right)^{2}$$

Apply the distributive property by multiplying each term of 4x+4 by each term of x-2.

$$4x^{2}-8x+4x-8=\left(-x+7\right)^{2}$$

Combine -8x and 4x to get -4x.

$$4x^{2}-4x-8=\left(-x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+7\right)^{2}.

$$4x^{2}-4x-8=x^{2}-14x+49$$

Subtract x^{2} from both sides.

$$4x^{2}-4x-8-x^{2}=-14x+49$$

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

$$3x^{2}-4x-8=-14x+49$$

Add 14x to both sides.

$$3x^{2}-4x-8+14x=49$$

Combine -4x and 14x to get 10x.

$$3x^{2}+10x-8=49$$

Subtract 49 from both sides.

$$3x^{2}+10x-8-49=0$$

Subtract 49 from -8 to get -57.

$$3x^{2}+10x-57=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-57. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\left(-57\right)=-171$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -171.

$$-1,171 -3,57 -9,19$$

Calculate the sum for each pair.

$$-1+171=170 -3+57=54 -9+19=10$$

The solution is the pair that gives sum 10.

$$a=-9 b=19$$

Rewrite 3x^{2}+10x-57 as \left(3x^{2}-9x\right)+\left(19x-57\right).

$$\left(3x^{2}-9x\right)+\left(19x-57\right)$$

Factor out 3x in the first and 19 in the second group.

$$3x\left(x-3\right)+19\left(x-3\right)$$

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

$$\left(x-3\right)\left(3x+19\right)$$

To find equation solutions, solve x-3=0 and 3x+19=0.

$$x=3 x=-\frac{19}{3}$$

Substitute -\frac{19}{3} for x in the equation \sqrt{x+1}+\sqrt{x-2}=\sqrt{x+6}. The expression \sqrt{-\frac{19}{3}+1} is undefined because the radicand cannot be negative.

$$\sqrt{-\frac{19}{3}+1}+\sqrt{-\frac{19}{3}-2}=\sqrt{-\frac{19}{3}+6}$$

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

$$\sqrt{3+1}+\sqrt{3-2}=\sqrt{3+6}$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

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

$$x=3$$