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

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

Solve for x

$$x=1$$

Solution Steps

Subtract -\sqrt{3x-2} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 2 from 1 to get -1.

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

Subtract -1+3x from both sides of the equation.

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

To find the opposite of -1+3x, find the opposite of each term.

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

Add 3 and 1 to get 4.

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

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

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

Square both sides of the equation.

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

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

$$4x^{2}-16x+16=\left(2\sqrt{3x-2}\right)^{2}$$

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

$$4x^{2}-16x+16=2^{2}\left(\sqrt{3x-2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-16x+16=4\left(\sqrt{3x-2}\right)^{2}$$

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

$$4x^{2}-16x+16=4\left(3x-2\right)$$

Use the distributive property to multiply 4 by 3x-2.

$$4x^{2}-16x+16=12x-8$$

Subtract 12x from both sides.

$$4x^{2}-16x+16-12x=-8$$

Combine -16x and -12x to get -28x.

$$4x^{2}-28x+16=-8$$

Add 8 to both sides.

$$4x^{2}-28x+16+8=0$$

Add 16 and 8 to get 24.

$$4x^{2}-28x+24=0$$

Divide both sides by 4.

$$x^{2}-7x+6=0$$

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.

$$a+b=-7 ab=1\times 6=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

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

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite x^{2}-7x+6 as \left(x^{2}-6x\right)+\left(-x+6\right).

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

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

$$x\left(x-6\right)-\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-1\right)$$

To find equation solutions, solve x-6=0 and x-1=0.

$$x=6 x=1$$

Substitute 6 for x in the equation \sqrt{x+3}-\sqrt{3x-2}=1.

$$\sqrt{6+3}-\sqrt{3\times 6-2}=1$$

Simplify. The value x=6 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

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

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

$$x=1$$