Solve -8/t^2-6t+t/t-6=1/t

$$\frac { - 8 } { t ^ { 2 } - 6 t } + \frac { t } { t - 6 } = \frac { 1 } { t }$$

Solve for t

$$t=-1$$

$$t=2$$

Steps Using Factoring

Variable t cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by t\left(t-6\right), the least common multiple of t^{2}-6t,t-6,t.

$$-8+tt=t-6$$

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

$$-8+t^{2}=t-6$$

Subtract t from both sides.

$$-8+t^{2}-t=-6$$

Add 6 to both sides.

$$-8+t^{2}-t+6=0$$

Add -8 and 6 to get -2.

$$-2+t^{2}-t=0$$

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

$$t^{2}-t-2=0$$

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

$$a+b=-1 ab=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

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

$$\left(t-2\right)\left(t+1\right)$$

To find equation solutions, solve t-2=0 and t+1=0.

$$t=2 t=-1$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by t\left(t-6\right), the least common multiple of t^{2}-6t,t-6,t.

$$-8+tt=t-6$$

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

$$-8+t^{2}=t-6$$

Subtract t from both sides.

$$-8+t^{2}-t=-6$$

Add 6 to both sides.

$$-8+t^{2}-t+6=0$$

Add -8 and 6 to get -2.

$$-2+t^{2}-t=0$$

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

$$t^{2}-t-2=0$$

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

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

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

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

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

Factor out t in t^{2}-2t.

$$t\left(t-2\right)+t-2$$

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

$$\left(t-2\right)\left(t+1\right)$$

To find equation solutions, solve t-2=0 and t+1=0.

$$t=2 t=-1$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by t\left(t-6\right), the least common multiple of t^{2}-6t,t-6,t.

$$-8+tt=t-6$$

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

$$-8+t^{2}=t-6$$

Subtract t from both sides.

$$-8+t^{2}-t=-6$$

Add 6 to both sides.

$$-8+t^{2}-t+6=0$$

Add -8 and 6 to get -2.

$$-2+t^{2}-t=0$$

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.

$$t^{2}-t-2=0$$

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

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

Multiply -4 times -2.

$$t=\frac{-\left(-1\right)±\sqrt{1+8}}{2}$$

Add 1 to 8.

$$t=\frac{-\left(-1\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$t=\frac{-\left(-1\right)±3}{2}$$

The opposite of -1 is 1.

$$t=\frac{1±3}{2}$$

Now solve the equation t=\frac{1±3}{2} when ± is plus. Add 1 to 3.

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

Divide 4 by 2.

$$t=2$$

Now solve the equation t=\frac{1±3}{2} when ± is minus. Subtract 3 from 1.

$$t=-\frac{2}{2}$$

Divide -2 by 2.

$$t=-1$$

The equation is now solved.

$$t=2 t=-1$$

Steps for Completing the Square

Variable t cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by t\left(t-6\right), the least common multiple of t^{2}-6t,t-6,t.

$$-8+tt=t-6$$

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

$$-8+t^{2}=t-6$$

Subtract t from both sides.

$$-8+t^{2}-t=-6$$

Add 8 to both sides.

$$t^{2}-t=-6+8$$

Add -6 and 8 to get 2.

$$t^{2}-t=2$$

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.

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

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

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

Add 2 to \frac{1}{4}.

$$t^{2}-t+\frac{1}{4}=\frac{9}{4}$$

Factor t^{2}-t+\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}.

$$\left(t-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$t-\frac{1}{2}=\frac{3}{2} t-\frac{1}{2}=-\frac{3}{2}$$

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

$$t=2 t=-1$$