Solve t+9/t=-10

$$t + \frac { 9 } { t } = - 10$$

Solve for t

$$t=-9$$

$$t=-1$$

Steps Using Factoring

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

$$tt+9=-10t$$

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

$$t^{2}+9=-10t$$

Add 10t to both sides.

$$t^{2}+9+10t=0$$

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

$$t^{2}+10t+9=0$$

To solve the equation, factor t^{2}+10t+9 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.

$$a+b=10 ab=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

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

$$\left(t+1\right)\left(t+9\right)$$

To find equation solutions, solve t+1=0 and t+9=0.

$$t=-1 t=-9$$

Steps Using Factoring By Grouping

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

$$tt+9=-10t$$

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

$$t^{2}+9=-10t$$

Add 10t to both sides.

$$t^{2}+9+10t=0$$

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

$$t^{2}+10t+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+9. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 9=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

Rewrite t^{2}+10t+9 as \left(t^{2}+t\right)+\left(9t+9\right).

$$\left(t^{2}+t\right)+\left(9t+9\right)$$

Factor out t in the first and 9 in the second group.

$$t\left(t+1\right)+9\left(t+1\right)$$

Factor out common term t+1 by using distributive property.

$$\left(t+1\right)\left(t+9\right)$$

To find equation solutions, solve t+1=0 and t+9=0.

$$t=-1 t=-9$$

Steps Using the Quadratic Formula

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

$$tt+9=-10t$$

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

$$t^{2}+9=-10t$$

Add 10t to both sides.

$$t^{2}+9+10t=0$$

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.

$$t^{2}+10t+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-10±\sqrt{10^{2}-4\times 9}}{2}$$

Square 10.

$$t=\frac{-10±\sqrt{100-4\times 9}}{2}$$

Multiply -4 times 9.

$$t=\frac{-10±\sqrt{100-36}}{2}$$

Add 100 to -36.

$$t=\frac{-10±\sqrt{64}}{2}$$

Take the square root of 64.

$$t=\frac{-10±8}{2}$$

Now solve the equation t=\frac{-10±8}{2} when ± is plus. Add -10 to 8.

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

Divide -2 by 2.

$$t=-1$$

Now solve the equation t=\frac{-10±8}{2} when ± is minus. Subtract 8 from -10.

$$t=-\frac{18}{2}$$

Divide -18 by 2.

$$t=-9$$

The equation is now solved.

$$t=-1 t=-9$$

Steps for Completing the Square

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

$$tt+9=-10t$$

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

$$t^{2}+9=-10t$$

Add 10t to both sides.

$$t^{2}+9+10t=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$t^{2}+10t=-9$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+10t+5^{2}=-9+5^{2}$$

Square 5.

$$t^{2}+10t+25=-9+25$$

Add -9 to 25.

$$t^{2}+10t+25=16$$

Factor t^{2}+10t+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+5\right)^{2}}=\sqrt{16}$$

Simplify.

$$t+5=4 t+5=-4$$

Subtract 5 from both sides of the equation.

$$t=-1 t=-9$$