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

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

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Combine 2x and x to get 3x.

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

Subtract 2 from -1 to get -3.

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

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

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

Subtract 3x-3 from both sides of the equation.

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

To find the opposite of 3x-3, find the opposite of each term.

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

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

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

Add 1 and 3 to get 4.

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

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

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

Use the distributive property to multiply 4 by 2x-1.

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

Apply the distributive property by multiplying each term of 8x-4 by each term of x-2.

$$8x^{2}-16x-4x+8=\left(-2x+4\right)^{2}$$

Combine -16x and -4x to get -20x.

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

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

$$8x^{2}-20x+8=4x^{2}-16x+16$$

Subtract 4x^{2} from both sides.

$$8x^{2}-20x+8-4x^{2}=-16x+16$$

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

$$4x^{2}-20x+8=-16x+16$$

Add 16x to both sides.

$$4x^{2}-20x+8+16x=16$$

Combine -20x and 16x to get -4x.

$$4x^{2}-4x+8=16$$

Subtract 16 from both sides.

$$4x^{2}-4x+8-16=0$$

Subtract 16 from 8 to get -8.

$$4x^{2}-4x-8=0$$

Divide both sides by 4.

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

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

$$a+b=-1 ab=1\left(-2\right)=-2$$

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. The only such pair is the system solution.

$$a=-2 b=1$$

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

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

Factor out x in x^{2}-2x.

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

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

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

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

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

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

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

Simplify. The value x=2 satisfies the equation.

$$3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

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

$$x=2$$