Solve sqrt{2x^2-2x+1}-2x+3=0

$$\sqrt{ 2 { x }^{ 2 } -2x+1 } -2x+3=0$$

Solve for x

$$x=4$$

Solution Steps

Subtract -2x+3 from both sides of the equation.

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

To find the opposite of -2x+3, find the opposite of each term.

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

Square both sides of the equation.

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

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

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

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

$$2x^{2}-2x+1=4x^{2}-12x+9$$

Subtract 4x^{2} from both sides.

$$2x^{2}-2x+1-4x^{2}=-12x+9$$

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

$$-2x^{2}-2x+1=-12x+9$$

Add 12x to both sides.

$$-2x^{2}-2x+1+12x=9$$

Combine -2x and 12x to get 10x.

$$-2x^{2}+10x+1=9$$

Subtract 9 from both sides.

$$-2x^{2}+10x+1-9=0$$

Subtract 9 from 1 to get -8.

$$-2x^{2}+10x-8=0$$

Divide both sides by 2.

$$-x^{2}+5x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=4 b=1$$

Rewrite -x^{2}+5x-4 as \left(-x^{2}+4x\right)+\left(x-4\right).

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

Factor out -x in -x^{2}+4x.

$$-x\left(x-4\right)+x-4$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x+1\right)$$

To find equation solutions, solve x-4=0 and -x+1=0.

$$x=4 x=1$$

Substitute 4 for x in the equation \sqrt{2x^{2}-2x+1}-2x+3=0.

$$\sqrt{2\times 4^{2}-2\times 4+1}-2\times 4+3=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Substitute 1 for x in the equation \sqrt{2x^{2}-2x+1}-2x+3=0.

$$\sqrt{2\times 1^{2}-2+1}-2+3=0$$

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

$$2=0$$

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

$$x=4$$