Solve m-6/m=-1

$$m - \frac { 6 } { m } = - 1$$

Solve for m

$$m=-3$$

$$m=2$$

Steps Using Factoring

Variable m cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by m.

$$mm-6=-m$$

Multiply m and m to get m^{2}.

$$m^{2}-6=-m$$

Add m to both sides.

$$m^{2}-6+m=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$m^{2}+m-6=0$$

To solve the equation, factor m^{2}+m-6 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.

$$a+b=1 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite factored expression \left(m+a\right)\left(m+b\right) using the obtained values.

$$\left(m-2\right)\left(m+3\right)$$

To find equation solutions, solve m-2=0 and m+3=0.

$$m=2 m=-3$$

Steps Using Factoring By Grouping

Variable m cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by m.

$$mm-6=-m$$

Multiply m and m to get m^{2}.

$$m^{2}-6=-m$$

Add m to both sides.

$$m^{2}-6+m=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$m^{2}+m-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite m^{2}+m-6 as \left(m^{2}-2m\right)+\left(3m-6\right).

$$\left(m^{2}-2m\right)+\left(3m-6\right)$$

Factor out m in the first and 3 in the second group.

$$m\left(m-2\right)+3\left(m-2\right)$$

Factor out common term m-2 by using distributive property.

$$\left(m-2\right)\left(m+3\right)$$

To find equation solutions, solve m-2=0 and m+3=0.

$$m=2 m=-3$$

Steps Using the Quadratic Formula

Variable m cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by m.

$$mm-6=-m$$

Multiply m and m to get m^{2}.

$$m^{2}-6=-m$$

Add m to both sides.

$$m^{2}-6+m=0$$

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.

$$m^{2}+m-6=0$$

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

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

Square 1.

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

Multiply -4 times -6.

$$m=\frac{-1±\sqrt{1+24}}{2}$$

Add 1 to 24.

$$m=\frac{-1±\sqrt{25}}{2}$$

Take the square root of 25.

$$m=\frac{-1±5}{2}$$

Now solve the equation m=\frac{-1±5}{2} when ± is plus. Add -1 to 5.

$$m=\frac{4}{2}$$

Divide 4 by 2.

$$m=2$$

Now solve the equation m=\frac{-1±5}{2} when ± is minus. Subtract 5 from -1.

$$m=-\frac{6}{2}$$

Divide -6 by 2.

$$m=-3$$

The equation is now solved.

$$m=2 m=-3$$

Steps for Completing the Square

Variable m cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by m.

$$mm-6=-m$$

Multiply m and m to get m^{2}.

$$m^{2}-6=-m$$

Add m to both sides.

$$m^{2}-6+m=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$m^{2}+m=6$$

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.

$$m^{2}+m+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

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

$$m^{2}+m+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$m^{2}+m+\frac{1}{4}=\frac{25}{4}$$

Factor m^{2}+m+\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}.

$$\left(m+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$m+\frac{1}{2}=\frac{5}{2} m+\frac{1}{2}=-\frac{5}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$m=2 m=-3$$