Solve {l}{(-ve)*(-ve)=(+ve)}{text{Multiply}(i)-8text{and}-7(text{ii})9}

$$\left. \begin{array} { l } { ( - v e ) \times ( - v e ) = ( + v e ) } \\ { \text { Multiply } ( i ) - 8 \text { and } - 7 ( \text { ii } ) 9 } \end{array} \right.$$

Solve for v

$$v=\frac{1}{e}\approx 0.367879441$$

$$v=0$$

Steps Using Factoring

Cancel out e on both sides.

$$\left(-v\right)\left(-v\right)e=v$$

Multiply -v and -v to get \left(-v\right)^{2}.

$$\left(-v\right)^{2}e=v$$

Calculate -v to the power of 2 and get v^{2}.

$$v^{2}e=v$$

Subtract v from both sides.

$$v^{2}e-v=0$$

Factor out v.

$$v\left(ve-1\right)=0$$

To find equation solutions, solve v=0 and ve-1=0.

$$v=0 v=\frac{1}{e}$$

Steps Using the Quadratic Formula

Cancel out e on both sides.

$$\left(-v\right)\left(-v\right)e=v$$

Multiply -v and -v to get \left(-v\right)^{2}.

$$\left(-v\right)^{2}e=v$$

Calculate -v to the power of 2 and get v^{2}.

$$v^{2}e=v$$

Subtract v from both sides.

$$v^{2}e-v=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$ev^{2}-v=0$$

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

$$v=\frac{-\left(-1\right)±\sqrt{1}}{2e}$$

Take the square root of 1.

$$v=\frac{-\left(-1\right)±1}{2e}$$

The opposite of -1 is 1.

$$v=\frac{1±1}{2e}$$

Now solve the equation v=\frac{1±1}{2e} when ± is plus. Add 1 to 1.

$$v=\frac{2}{2e}$$

Divide 2 by 2e.

$$v=\frac{1}{e}$$

Now solve the equation v=\frac{1±1}{2e} when ± is minus. Subtract 1 from 1.

$$v=\frac{0}{2e}$$

Divide 0 by 2e.

$$v=0$$

The equation is now solved.

$$v=\frac{1}{e} v=0$$

Steps for Completing the Square

Cancel out e on both sides.

$$\left(-v\right)\left(-v\right)e=v$$

Multiply -v and -v to get \left(-v\right)^{2}.

$$\left(-v\right)^{2}e=v$$

Calculate -v to the power of 2 and get v^{2}.

$$v^{2}e=v$$

Subtract v from both sides.

$$v^{2}e-v=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$ev^{2}-v=0$$

Divide both sides by e.

$$\frac{ev^{2}-v}{e}=\frac{0}{e}$$

Dividing by e undoes the multiplication by e.

$$v^{2}+\left(-\frac{1}{e}\right)v=\frac{0}{e}$$

Divide 0 by e.

$$v^{2}+\left(-\frac{1}{e}\right)v=0$$

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

$$v^{2}+\left(-\frac{1}{e}\right)v+\left(-\frac{1}{2e}\right)^{2}=\left(-\frac{1}{2e}\right)^{2}$$

Square -\frac{1}{2e}.

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

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

$$\left(v-\frac{1}{2e}\right)^{2}=\frac{1}{4e^{2}}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-\frac{1}{2e}\right)^{2}}=\sqrt{\frac{1}{4e^{2}}}$$

Simplify.

$$v-\frac{1}{2e}=\frac{1}{2e} v-\frac{1}{2e}=-\frac{1}{2e}$$

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

$$v=\frac{1}{e} v=0$$