Solve (x^2-7x)-(x-x^2)=x

$$( x ^ { 2 } - 7 x ) - ( x - x ^ { 2 } ) = x$$

Solve for x

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

$$x=0$$

Steps Using Factoring

To find the opposite of x-x^{2}, find the opposite of each term.

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

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

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

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

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

Subtract x from both sides.

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

Combine -8x and -x to get -9x.

$$2x^{2}-9x=0$$

Factor out x.

$$x\left(2x-9\right)=0$$

To find equation solutions, solve x=0 and 2x-9=0.

$$x=0 x=\frac{9}{2}$$

Steps Using the Quadratic Formula

To find the opposite of x-x^{2}, find the opposite of each term.

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

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

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

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

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

Subtract x from both sides.

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

Combine -8x and -x to get -9x.

$$2x^{2}-9x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}}}{2\times 2}$$

Take the square root of \left(-9\right)^{2}.

$$x=\frac{-\left(-9\right)±9}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±9}{4}$$

Now solve the equation x=\frac{9±9}{4} when ± is plus. Add 9 to 9.

$$x=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{9±9}{4} when ± is minus. Subtract 9 from 9.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=\frac{9}{2} x=0$$

Steps for Completing the Square

To find the opposite of x-x^{2}, find the opposite of each term.

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

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

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

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

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

Subtract x from both sides.

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

Combine -8x and -x to get -9x.

$$2x^{2}-9x=0$$

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=\frac{0}{2}$$

Divide 0 by 2.

$$x^{2}-\frac{9}{2}x=0$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{9}{4} x-\frac{9}{4}=-\frac{9}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{9}{2} x=0$$