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

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

Solve for x

$$x=2$$

Solution Steps

Subtract \sqrt{x-1} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2x+5}-\sqrt{x-1}\right)^{2}.

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

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

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

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

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

Combine 2x and x to get 3x.

$$x+2=3x+5-2\sqrt{2x+5}\sqrt{x-1}-1$$

Subtract 1 from 5 to get 4.

$$x+2=3x+4-2\sqrt{2x+5}\sqrt{x-1}$$

Subtract 3x+4 from both sides of the equation.

$$x+2-\left(3x+4\right)=-2\sqrt{2x+5}\sqrt{x-1}$$

To find the opposite of 3x+4, find the opposite of each term.

$$x+2-3x-4=-2\sqrt{2x+5}\sqrt{x-1}$$

Combine x and -3x to get -2x.

$$-2x+2-4=-2\sqrt{2x+5}\sqrt{x-1}$$

Subtract 4 from 2 to get -2.

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

Square both sides of the equation.

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

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

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

Expand \left(-2\sqrt{2x+5}\sqrt{x-1}\right)^{2}.

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

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

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

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

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

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

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

Use the distributive property to multiply 4 by 2x+5.

$$4x^{2}+8x+4=\left(8x+20\right)\left(x-1\right)$$

Apply the distributive property by multiplying each term of 8x+20 by each term of x-1.

$$4x^{2}+8x+4=8x^{2}-8x+20x-20$$

Combine -8x and 20x to get 12x.

$$4x^{2}+8x+4=8x^{2}+12x-20$$

Subtract 8x^{2} from both sides.

$$4x^{2}+8x+4-8x^{2}=12x-20$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$-4x^{2}+8x+4=12x-20$$

Subtract 12x from both sides.

$$-4x^{2}+8x+4-12x=-20$$

Combine 8x and -12x to get -4x.

$$-4x^{2}-4x+4=-20$$

Add 20 to both sides.

$$-4x^{2}-4x+4+20=0$$

Add 4 and 20 to get 24.

$$-4x^{2}-4x+24=0$$

Divide both sides by 4.

$$-x^{2}-x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-6=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

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

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

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

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

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

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

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

$$x=2 x=-3$$

Substitute -3 for x in the equation \sqrt{x+2}+\sqrt{x-1}=\sqrt{2x+5}. The expression \sqrt{-3+2} is undefined because the radicand cannot be negative.

$$\sqrt{-3+2}+\sqrt{-3-1}=\sqrt{2\left(-3\right)+5}$$

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

$$\sqrt{2+2}+\sqrt{2-1}=\sqrt{2\times 2+5}$$

Simplify. The value x=2 satisfies the equation.

$$3=3$$

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

$$x=2$$