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

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

Solve for x

$$x=1$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

$$3x-2=9-6\sqrt{x+3}+\left(\sqrt{x+3}\right)^{2}$$

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

$$3x-2=9-6\sqrt{x+3}+x+3$$

Add 9 and 3 to get 12.

$$3x-2=12-6\sqrt{x+3}+x$$

Subtract 12+x from both sides of the equation.

$$3x-2-\left(12+x\right)=-6\sqrt{x+3}$$

To find the opposite of 12+x, find the opposite of each term.

$$3x-2-12-x=-6\sqrt{x+3}$$

Subtract 12 from -2 to get -14.

$$3x-14-x=-6\sqrt{x+3}$$

Combine 3x and -x to get 2x.

$$2x-14=-6\sqrt{x+3}$$

Square both sides of the equation.

$$\left(2x-14\right)^{2}=\left(-6\sqrt{x+3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-14\right)^{2}.

$$4x^{2}-56x+196=\left(-6\sqrt{x+3}\right)^{2}$$

Expand \left(-6\sqrt{x+3}\right)^{2}.

$$4x^{2}-56x+196=\left(-6\right)^{2}\left(\sqrt{x+3}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$4x^{2}-56x+196=36\left(\sqrt{x+3}\right)^{2}$$

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

$$4x^{2}-56x+196=36\left(x+3\right)$$

Use the distributive property to multiply 36 by x+3.

$$4x^{2}-56x+196=36x+108$$

Subtract 36x from both sides.

$$4x^{2}-56x+196-36x=108$$

Combine -56x and -36x to get -92x.

$$4x^{2}-92x+196=108$$

Subtract 108 from both sides.

$$4x^{2}-92x+196-108=0$$

Subtract 108 from 196 to get 88.

$$4x^{2}-92x+88=0$$

Divide both sides by 4.

$$x^{2}-23x+22=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+22. To find a and b, set up a system to be solved.

$$a+b=-23 ab=1\times 22=22$$

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 22.

$$-1,-22 -2,-11$$

Calculate the sum for each pair.

$$-1-22=-23 -2-11=-13$$

The solution is the pair that gives sum -23.

$$a=-22 b=-1$$

Rewrite x^{2}-23x+22 as \left(x^{2}-22x\right)+\left(-x+22\right).

$$\left(x^{2}-22x\right)+\left(-x+22\right)$$

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

$$x\left(x-22\right)-\left(x-22\right)$$

Factor out common term x-22 by using distributive property.

$$\left(x-22\right)\left(x-1\right)$$

To find equation solutions, solve x-22=0 and x-1=0.

$$x=22 x=1$$

Substitute 22 for x in the equation \sqrt{3x-2}+\sqrt{x+3}=3.

$$\sqrt{3\times 22-2}+\sqrt{22+3}=3$$

Simplify. The value x=22 does not satisfy the equation.

$$13=3$$

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

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

Simplify. The value x=1 satisfies the equation.

$$3=3$$

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

$$x=1$$