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

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

Solve for x

$$x=5$$

$$x=1$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 1 from 4 to get 3.

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

Subtract 3+x from both sides of the equation.

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

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

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

Subtract 3 from 1 to get -2.

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

Combine 3x and -x to get 2x.

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

Square both sides of the equation.

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

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

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

Expand \left(4\sqrt{x-1}\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$4x^{2}-8x+4=16\left(\sqrt{x-1}\right)^{2}$$

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

$$4x^{2}-8x+4=16\left(x-1\right)$$

Use the distributive property to multiply 16 by x-1.

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

Subtract 16x from both sides.

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

Combine -8x and -16x to get -24x.

$$4x^{2}-24x+4=-16$$

Add 16 to both sides.

$$4x^{2}-24x+4+16=0$$

Add 4 and 16 to get 20.

$$4x^{2}-24x+20=0$$

Divide both sides by 4.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 5=5$$

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 x^{2}-6x+5 as \left(x^{2}-5x\right)+\left(-x+5\right).

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

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

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

Factor out common term x-5 by using distributive property.

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

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{x-1}=2.

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

Simplify. The value x=5 satisfies the equation.

$$2=2$$

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

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

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

$$x=5 x=1$$