Solve sqrt{3x-5}=1+sqrt{x-2}

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

Solve for x

$$x=3$$

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x-5}\right)^{2}=\left(1+\sqrt{x-2}\right)^{2}$$

Calculate \sqrt{3x-5} to the power of 2 and get 3x-5.

$$3x-5=\left(1+\sqrt{x-2}\right)^{2}$$

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

$$3x-5=1+2\sqrt{x-2}+\left(\sqrt{x-2}\right)^{2}$$

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

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

Subtract 2 from 1 to get -1.

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

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

$$3x-5-\left(-1+x\right)=2\sqrt{x-2}$$

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

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

Add -5 and 1 to get -4.

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

Combine 3x and -x to get 2x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

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

Subtract 4x from both sides.

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

Combine -16x and -4x to get -20x.

$$4x^{2}-20x+16=-8$$

Add 8 to both sides.

$$4x^{2}-20x+16+8=0$$

Add 16 and 8 to get 24.

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

Divide both sides by 4.

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

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.

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

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 -5.

$$a=-3 b=-2$$

Rewrite x^{2}-5x+6 as \left(x^{2}-3x\right)+\left(-2x+6\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-3\right)-2\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-2\right)$$

To find equation solutions, solve x-3=0 and x-2=0.

$$x=3 x=2$$

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

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

Simplify. The value x=3 satisfies the equation.

$$2=2$$

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

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

Simplify. The value x=2 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{3x-5}=\sqrt{x-2}+1.

$$x=3 x=2$$