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

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

Solve for x

$$x=6$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add 3 and 1 to get 4.

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

Subtract x+4 from both sides of the equation.

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

To find the opposite of x+4, find the opposite of each term.

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

Combine 3x and -x to get 2x.

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

Subtract 4 from -2 to get -6.

$$2x-6=2\sqrt{x+3}$$

Square both sides of the equation.

$$\left(2x-6\right)^{2}=\left(2\sqrt{x+3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-6\right)^{2}.

$$4x^{2}-24x+36=\left(2\sqrt{x+3}\right)^{2}$$

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

$$4x^{2}-24x+36=2^{2}\left(\sqrt{x+3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-24x+36=4\left(\sqrt{x+3}\right)^{2}$$

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

$$4x^{2}-24x+36=4\left(x+3\right)$$

Use the distributive property to multiply 4 by x+3.

$$4x^{2}-24x+36=4x+12$$

Subtract 4x from both sides.

$$4x^{2}-24x+36-4x=12$$

Combine -24x and -4x to get -28x.

$$4x^{2}-28x+36=12$$

Subtract 12 from both sides.

$$4x^{2}-28x+36-12=0$$

Subtract 12 from 36 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{3x-2}=\sqrt{x+3}+1.

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

Simplify. The value x=6 satisfies the equation.

$$4=4$$

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

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

Simplify. The value x=1 does not satisfy the equation.

$$1=3$$

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

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

Simplify. The value x=6 satisfies the equation.

$$4=4$$

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

$$x=6$$