Solve (x-3)^2+1-2(x-3)^2/(x-3)^2=0

$$\frac { ( x - 3 ) ^ { 2 } + 1 - 2 ( x - 3 ) ^ { 2 } } { ( x - 3 ) ^ { 2 } } = 0$$

Solve for x

$$x=2$$

$$x=4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}.

$$\left(x-3\right)^{2}+1-2\left(x-3\right)^{2}=0$$

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

$$x^{2}-6x+9+1-2\left(x-3\right)^{2}=0$$

Add 9 and 1 to get 10.

$$x^{2}-6x+10-2\left(x-3\right)^{2}=0$$

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

$$x^{2}-6x+10-2\left(x^{2}-6x+9\right)=0$$

Use the distributive property to multiply -2 by x^{2}-6x+9.

$$x^{2}-6x+10-2x^{2}+12x-18=0$$

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

$$-x^{2}-6x+10+12x-18=0$$

Combine -6x and 12x to get 6x.

$$-x^{2}+6x+10-18=0$$

Subtract 18 from 10 to get -8.

$$-x^{2}+6x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-8\right)=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=4 b=2$$

Rewrite -x^{2}+6x-8 as \left(-x^{2}+4x\right)+\left(2x-8\right).

$$\left(-x^{2}+4x\right)+\left(2x-8\right)$$

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

$$-x\left(x-4\right)+2\left(x-4\right)$$

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

$$\left(x-4\right)\left(-x+2\right)$$

To find equation solutions, solve x-4=0 and -x+2=0.

$$x=4 x=2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}.

$$\left(x-3\right)^{2}+1-2\left(x-3\right)^{2}=0$$

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

$$x^{2}-6x+9+1-2\left(x-3\right)^{2}=0$$

Add 9 and 1 to get 10.

$$x^{2}-6x+10-2\left(x-3\right)^{2}=0$$

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

$$x^{2}-6x+10-2\left(x^{2}-6x+9\right)=0$$

Use the distributive property to multiply -2 by x^{2}-6x+9.

$$x^{2}-6x+10-2x^{2}+12x-18=0$$

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

$$-x^{2}-6x+10+12x-18=0$$

Combine -6x and 12x to get 6x.

$$-x^{2}+6x+10-18=0$$

Subtract 18 from 10 to get -8.

$$-x^{2}+6x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-8\right)}}{2\left(-1\right)}$$

Multiply 4 times -8.

$$x=\frac{-6±\sqrt{36-32}}{2\left(-1\right)}$$

Add 36 to -32.

$$x=\frac{-6±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$x=\frac{-6±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±2}{-2}$$

Now solve the equation x=\frac{-6±2}{-2} when ± is plus. Add -6 to 2.

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

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-6±2}{-2} when ± is minus. Subtract 2 from -6.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=2 x=4$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}.

$$\left(x-3\right)^{2}+1-2\left(x-3\right)^{2}=0$$

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

$$x^{2}-6x+9+1-2\left(x-3\right)^{2}=0$$

Add 9 and 1 to get 10.

$$x^{2}-6x+10-2\left(x-3\right)^{2}=0$$

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

$$x^{2}-6x+10-2\left(x^{2}-6x+9\right)=0$$

Use the distributive property to multiply -2 by x^{2}-6x+9.

$$x^{2}-6x+10-2x^{2}+12x-18=0$$

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

$$-x^{2}-6x+10+12x-18=0$$

Combine -6x and 12x to get 6x.

$$-x^{2}+6x+10-18=0$$

Subtract 18 from 10 to get -8.

$$-x^{2}+6x-8=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$-x^{2}+6x=8$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{8}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{8}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{8}{-1}$$

Divide 8 by -1.

$$x^{2}-6x=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-8+9$$

Add -8 to 9.

$$x^{2}-6x+9=1$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=1 x-3=-1$$

Add 3 to both sides of the equation.

$$x=4 x=2$$