Solve (x-2)(x-1)=-2(1-x)^2

$$( x - 2 ) ( x - 1 ) = - 2 ( 1 - x ) ^ { 2 }$$

Solve for x

$$x=1$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x-2 by x-1 and combine like terms.

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

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

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

Use the distributive property to multiply -2 by 1-2x+x^{2}.

$$x^{2}-3x+2=-2+4x-2x^{2}$$

Subtract -2 from both sides.

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

The opposite of -2 is 2.

$$x^{2}-3x+2+2=4x-2x^{2}$$

Combine 2 and 2 to get 2\times 2.

$$x^{2}-3x+2\times 2=4x-2x^{2}$$

Subtract 4x from both sides.

$$x^{2}-3x+2\times 2-4x=-2x^{2}$$

Multiply 2 and 2 to get 4.

$$x^{2}-3x+4-4x=-2x^{2}$$

Combine -3x and -4x to get -7x.

$$x^{2}-7x+4=-2x^{2}$$

Add 2x^{2} to both sides.

$$x^{2}-7x+4+2x^{2}=0$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-7x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-7 ab=3\times 4=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite 3x^{2}-7x+4 as \left(3x^{2}-4x\right)+\left(-3x+4\right).

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

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

$$x\left(3x-4\right)-\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x-1\right)$$

To find equation solutions, solve 3x-4=0 and x-1=0.

$$x=\frac{4}{3} x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-2 by x-1 and combine like terms.

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

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

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

Use the distributive property to multiply -2 by 1-2x+x^{2}.

$$x^{2}-3x+2=-2+4x-2x^{2}$$

Subtract -2 from both sides.

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

The opposite of -2 is 2.

$$x^{2}-3x+2+2=4x-2x^{2}$$

Combine 2 and 2 to get 2\times 2.

$$x^{2}-3x+2\times 2=4x-2x^{2}$$

Subtract 4x from both sides.

$$x^{2}-3x+2\times 2-4x=-2x^{2}$$

Multiply 2 and 2 to get 4.

$$x^{2}-3x+4-4x=-2x^{2}$$

Combine -3x and -4x to get -7x.

$$x^{2}-7x+4=-2x^{2}$$

Add 2x^{2} to both sides.

$$x^{2}-7x+4+2x^{2}=0$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-7x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 3}$$

Add 49 to -48.

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

Take the square root of 1.

$$x=\frac{-\left(-7\right)±1}{2\times 3}$$

The opposite of -7 is 7.

$$x=\frac{7±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{7±1}{6}$$

Now solve the equation x=\frac{7±1}{6} when ± is plus. Add 7 to 1.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{7±1}{6} when ± is minus. Subtract 1 from 7.

$$x=\frac{6}{6}$$

Divide 6 by 6.

$$x=1$$

The equation is now solved.

$$x=\frac{4}{3} x=1$$

Steps for Completing the Square

Use the distributive property to multiply x-2 by x-1 and combine like terms.

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

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

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

Use the distributive property to multiply -2 by 1-2x+x^{2}.

$$x^{2}-3x+2=-2+4x-2x^{2}$$

Subtract 4x from both sides.

$$x^{2}-3x+2-4x=-2-2x^{2}$$

Combine -3x and -4x to get -7x.

$$x^{2}-7x+2=-2-2x^{2}$$

Add 2x^{2} to both sides.

$$x^{2}-7x+2+2x^{2}=-2$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-7x+2=-2$$

Subtract 2 from both sides.

$$3x^{2}-7x=-2-2$$

Subtract 2 from -2 to get -4.

$$3x^{2}-7x=-4$$

Divide both sides by 3.

$$\frac{3x^{2}-7x}{3}=-\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{7}{3}x=-\frac{4}{3}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-\frac{4}{3}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{4}{3}+\frac{49}{36}$$

Add -\frac{4}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{1}{6} x-\frac{7}{6}=-\frac{1}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{4}{3} x=1$$