Solve sqrt{2x-3}+sqrt{x-1}=sqrt{3x-2}

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

Solve for x

$$x=2$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3x-2}-\sqrt{x-1}\right)^{2}.

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

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

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

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

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

Combine 3x and x to get 4x.

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

Subtract 1 from -2 to get -3.

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

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

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

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

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

Combine 2x and -4x to get -2x.

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

Add -3 and 3 to get 0.

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

Cancel out -2 on both sides.

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

Square both sides of the equation.

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

Expand \left(\sqrt{3x-2}\sqrt{x-1}\right)^{2}.

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

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

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

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

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

Apply the distributive property by multiplying each term of 3x-2 by each term of x-1.

$$x^{2}=3x^{2}-3x-2x+2$$

Combine -3x and -2x to get -5x.

$$x^{2}=3x^{2}-5x+2$$

Subtract 3x^{2} from both sides.

$$x^{2}-3x^{2}=-5x+2$$

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

$$-2x^{2}=-5x+2$$

Add 5x to both sides.

$$-2x^{2}+5x=2$$

Subtract 2 from both sides.

$$-2x^{2}+5x-2=0$$

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

$$a+b=5 ab=-2\left(-2\right)=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=4 b=1$$

Rewrite -2x^{2}+5x-2 as \left(-2x^{2}+4x\right)+\left(x-2\right).

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

Factor out 2x in the first and -1 in the second group.

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

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

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

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

$$x=2 x=\frac{1}{2}$$

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

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

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

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

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Equation \sqrt{2x-3}=\sqrt{3x-2}-\sqrt{x-1} has a unique solution.

$$x=2$$