Solve 7(x^2-x)/3+2/3=x-x^2/3

$$\frac { 7 ( x ^ { 2 } - x ) } { 3 } + \frac { 2 } { 3 } = \frac { x - x ^ { 2 } } { 3 }$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 3.

$$7\left(x^{2}-x\right)+2=x-x^{2}$$

Use the distributive property to multiply 7 by x^{2}-x.

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

Subtract x from both sides.

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

Combine -7x and -x to get -8x.

$$7x^{2}-8x+2=-x^{2}$$

Add x^{2} to both sides.

$$7x^{2}-8x+2+x^{2}=0$$

Combine 7x^{2} and x^{2} to get 8x^{2}.

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

Divide both sides by 2.

$$4x^{2}-4x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=4\times 1=4$$

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 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 -4.

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

Rewrite 4x^{2}-4x+1 as \left(4x^{2}-2x\right)+\left(-2x+1\right).

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(2x-1\right)-\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(2x-1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve 2x-1=0.

$$x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$7\left(x^{2}-x\right)+2=x-x^{2}$$

Use the distributive property to multiply 7 by x^{2}-x.

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

Subtract x from both sides.

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

Combine -7x and -x to get -8x.

$$7x^{2}-8x+2=-x^{2}$$

Add x^{2} to both sides.

$$7x^{2}-8x+2+x^{2}=0$$

Combine 7x^{2} and x^{2} to get 8x^{2}.

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

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

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

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 8\times 2}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-8\right)±\sqrt{64-32\times 2}}{2\times 8}$$

Multiply -32 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-64}}{2\times 8}$$

Add 64 to -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2\times 8}$$

Take the square root of 0.

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

The opposite of -8 is 8.

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

Multiply 2 times 8.

$$x=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$7\left(x^{2}-x\right)+2=x-x^{2}$$

Use the distributive property to multiply 7 by x^{2}-x.

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

Subtract x from both sides.

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

Combine -7x and -x to get -8x.

$$7x^{2}-8x+2=-x^{2}$$

Add x^{2} to both sides.

$$7x^{2}-8x+2+x^{2}=0$$

Combine 7x^{2} and x^{2} to get 8x^{2}.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

Divide -8 by 8.

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

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

$$x^{2}-x=-\frac{1}{4}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{-1+1}{4}$$

Add -\frac{1}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=0$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{2}=0 x-\frac{1}{2}=0$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{1}{2} x=\frac{1}{2}$$

The equation is now solved. Solutions are the same.

$$x=\frac{1}{2}$$