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

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

Solve for x

$$x=1$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Add 4 and 3 to get 7.

$$2x-1=7-4\sqrt{3-2x}-2x$$

Subtract 7-2x from both sides of the equation.

$$2x-1-\left(7-2x\right)=-4\sqrt{3-2x}$$

To find the opposite of 7-2x, find the opposite of each term.

$$2x-1-7+2x=-4\sqrt{3-2x}$$

Subtract 7 from -1 to get -8.

$$2x-8+2x=-4\sqrt{3-2x}$$

Combine 2x and 2x to get 4x.

$$4x-8=-4\sqrt{3-2x}$$

Square both sides of the equation.

$$\left(4x-8\right)^{2}=\left(-4\sqrt{3-2x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-8\right)^{2}.

$$16x^{2}-64x+64=\left(-4\sqrt{3-2x}\right)^{2}$$

Expand \left(-4\sqrt{3-2x}\right)^{2}.

$$16x^{2}-64x+64=\left(-4\right)^{2}\left(\sqrt{3-2x}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$16x^{2}-64x+64=16\left(\sqrt{3-2x}\right)^{2}$$

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

$$16x^{2}-64x+64=16\left(3-2x\right)$$

Use the distributive property to multiply 16 by 3-2x.

$$16x^{2}-64x+64=48-32x$$

Subtract 48 from both sides.

$$16x^{2}-64x+64-48=-32x$$

Subtract 48 from 64 to get 16.

$$16x^{2}-64x+16=-32x$$

Add 32x to both sides.

$$16x^{2}-64x+16+32x=0$$

Combine -64x and 32x to get -32x.

$$16x^{2}-32x+16=0$$

Divide both sides by 16.

$$x^{2}-2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\times 1=1$$

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.

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

Rewrite x^{2}-2x+1 as \left(x^{2}-x\right)+\left(-x+1\right).

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

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

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

Factor out common term x-1 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve x-1=0.

$$x=1$$

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

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

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

$$x=1$$