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

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

Solve for x

$$x=5$$

$$x=1$$

Solution Steps

Subtract -\sqrt{2x-1} from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{3x+1} to the power of 2 and get 3x+1.

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

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

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

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

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

Subtract 1 from 1 to get 0.

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

Subtract 2x from both sides of the equation.

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

Combine 3x and -2x to get x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

$$x^{2}+2x+1=8x-4$$

Subtract 8x from both sides.

$$x^{2}+2x+1-8x=-4$$

Combine 2x and -8x to get -6x.

$$x^{2}-6x+1=-4$$

Add 4 to both sides.

$$x^{2}-6x+1+4=0$$

Add 1 and 4 to get 5.

$$x^{2}-6x+5=0$$

To solve the equation, factor x^{2}-6x+5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-5\right)\left(x-1\right)$$

To find equation solutions, solve x-5=0 and x-1=0.

$$x=5 x=1$$

Substitute 5 for x in the equation \sqrt{3x+1}-\sqrt{2x-1}=1.

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

Simplify. The value x=5 satisfies the equation.

$$1=1$$

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

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{3x+1}=\sqrt{2x-1}+1.

$$x=5 x=1$$