Solve 2m^2+m-3/m^2-m=1

$$\frac { 2 m ^ { 2 } + m - 3 } { m ^ { 2 } - m } = 1$$

Solve for m

$$m=-3$$

Steps Using Factoring

Variable m cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by m\left(m-1\right).

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

Use the distributive property to multiply m by m-1.

$$2m^{2}+m-3=m^{2}-m$$

Subtract m^{2} from both sides.

$$2m^{2}+m-3-m^{2}=-m$$

Combine 2m^{2} and -m^{2} to get m^{2}.

$$m^{2}+m-3=-m$$

Add m to both sides.

$$m^{2}+m-3+m=0$$

Combine m and m to get 2m.

$$m^{2}+2m-3=0$$

To solve the equation, factor m^{2}+2m-3 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.

$$a+b=2 ab=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

$$\left(m-1\right)\left(m+3\right)$$

To find equation solutions, solve m-1=0 and m+3=0.

$$m=1 m=-3$$

Variable m cannot be equal to 1.

$$m=-3$$

Steps Using Factoring By Grouping

Variable m cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by m\left(m-1\right).

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

Use the distributive property to multiply m by m-1.

$$2m^{2}+m-3=m^{2}-m$$

Subtract m^{2} from both sides.

$$2m^{2}+m-3-m^{2}=-m$$

Combine 2m^{2} and -m^{2} to get m^{2}.

$$m^{2}+m-3=-m$$

Add m to both sides.

$$m^{2}+m-3+m=0$$

Combine m and m to get 2m.

$$m^{2}+2m-3=0$$

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

$$a+b=2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

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

$$m\left(m-1\right)+3\left(m-1\right)$$

Factor out common term m-1 by using distributive property.

$$\left(m-1\right)\left(m+3\right)$$

To find equation solutions, solve m-1=0 and m+3=0.

$$m=1 m=-3$$

Variable m cannot be equal to 1.

$$m=-3$$

Steps Using the Quadratic Formula

Variable m cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by m\left(m-1\right).

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

Use the distributive property to multiply m by m-1.

$$2m^{2}+m-3=m^{2}-m$$

Subtract m^{2} from both sides.

$$2m^{2}+m-3-m^{2}=-m$$

Combine 2m^{2} and -m^{2} to get m^{2}.

$$m^{2}+m-3=-m$$

Add m to both sides.

$$m^{2}+m-3+m=0$$

Combine m and m to get 2m.

$$m^{2}+2m-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

$$m=\frac{-2±\sqrt{4-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$m=\frac{-2±\sqrt{4+12}}{2}$$

Add 4 to 12.

$$m=\frac{-2±\sqrt{16}}{2}$$

Take the square root of 16.

$$m=\frac{-2±4}{2}$$

Now solve the equation m=\frac{-2±4}{2} when ± is plus. Add -2 to 4.

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

Divide 2 by 2.

$$m=1$$

Now solve the equation m=\frac{-2±4}{2} when ± is minus. Subtract 4 from -2.

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

Divide -6 by 2.

$$m=-3$$

The equation is now solved.

$$m=1 m=-3$$

Variable m cannot be equal to 1.

$$m=-3$$

Steps for Completing the Square

Variable m cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by m\left(m-1\right).

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

Use the distributive property to multiply m by m-1.

$$2m^{2}+m-3=m^{2}-m$$

Subtract m^{2} from both sides.

$$2m^{2}+m-3-m^{2}=-m$$

Combine 2m^{2} and -m^{2} to get m^{2}.

$$m^{2}+m-3=-m$$

Add m to both sides.

$$m^{2}+m-3+m=0$$

Combine m and m to get 2m.

$$m^{2}+2m-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$m^{2}+2m=3$$

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

$$m^{2}+2m+1^{2}=3+1^{2}$$

Square 1.

$$m^{2}+2m+1=3+1$$

Add 3 to 1.

$$m^{2}+2m+1=4$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$m+1=2 m+1=-2$$

Subtract 1 from both sides of the equation.

$$m=1 m=-3$$

Variable m cannot be equal to 1.

$$m=-3$$