Solve t-4/t^2-3t=-2/t^2-9

$$\frac { t - 4 } { t ^ { 2 } - 3 t } = \frac { - 2 } { t ^ { 2 } - 9 }$$

Solve for t

$$t=-4$$

Steps Using Factoring

Variable t cannot be equal to any of the values -3,0,3 since division by zero is not defined. Multiply both sides of the equation by t\left(t-3\right)\left(t+3\right), the least common multiple of t^{2}-3t,t^{2}-9.

$$\left(t+3\right)\left(t-4\right)=t\left(-2\right)$$

Use the distributive property to multiply t+3 by t-4 and combine like terms.

$$t^{2}-t-12=t\left(-2\right)$$

Subtract t\left(-2\right) from both sides.

$$t^{2}-t-12-t\left(-2\right)=0$$

Combine -t and -t\left(-2\right) to get t.

$$t^{2}+t-12=0$$

To solve the equation, factor t^{2}+t-12 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.

$$a+b=1 ab=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

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

$$\left(t-3\right)\left(t+4\right)$$

To find equation solutions, solve t-3=0 and t+4=0.

$$t=3 t=-4$$

Variable t cannot be equal to 3.

$$t=-4$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values -3,0,3 since division by zero is not defined. Multiply both sides of the equation by t\left(t-3\right)\left(t+3\right), the least common multiple of t^{2}-3t,t^{2}-9.

$$\left(t+3\right)\left(t-4\right)=t\left(-2\right)$$

Use the distributive property to multiply t+3 by t-4 and combine like terms.

$$t^{2}-t-12=t\left(-2\right)$$

Subtract t\left(-2\right) from both sides.

$$t^{2}-t-12-t\left(-2\right)=0$$

Combine -t and -t\left(-2\right) to get t.

$$t^{2}+t-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-12. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-12\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite t^{2}+t-12 as \left(t^{2}-3t\right)+\left(4t-12\right).

$$\left(t^{2}-3t\right)+\left(4t-12\right)$$

Factor out t in the first and 4 in the second group.

$$t\left(t-3\right)+4\left(t-3\right)$$

Factor out common term t-3 by using distributive property.

$$\left(t-3\right)\left(t+4\right)$$

To find equation solutions, solve t-3=0 and t+4=0.

$$t=3 t=-4$$

Variable t cannot be equal to 3.

$$t=-4$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values -3,0,3 since division by zero is not defined. Multiply both sides of the equation by t\left(t-3\right)\left(t+3\right), the least common multiple of t^{2}-3t,t^{2}-9.

$$\left(t+3\right)\left(t-4\right)=t\left(-2\right)$$

Use the distributive property to multiply t+3 by t-4 and combine like terms.

$$t^{2}-t-12=t\left(-2\right)$$

Subtract t\left(-2\right) from both sides.

$$t^{2}-t-12-t\left(-2\right)=0$$

Combine -t and -t\left(-2\right) to get t.

$$t^{2}+t-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-12\right)}}{2}$$

Square 1.

$$t=\frac{-1±\sqrt{1-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$t=\frac{-1±\sqrt{1+48}}{2}$$

Add 1 to 48.

$$t=\frac{-1±\sqrt{49}}{2}$$

Take the square root of 49.

$$t=\frac{-1±7}{2}$$

Now solve the equation t=\frac{-1±7}{2} when ± is plus. Add -1 to 7.

$$t=\frac{6}{2}$$

Divide 6 by 2.

$$t=3$$

Now solve the equation t=\frac{-1±7}{2} when ± is minus. Subtract 7 from -1.

$$t=-\frac{8}{2}$$

Divide -8 by 2.

$$t=-4$$

The equation is now solved.

$$t=3 t=-4$$

Variable t cannot be equal to 3.

$$t=-4$$

Steps for Completing the Square

Variable t cannot be equal to any of the values -3,0,3 since division by zero is not defined. Multiply both sides of the equation by t\left(t-3\right)\left(t+3\right), the least common multiple of t^{2}-3t,t^{2}-9.

$$\left(t+3\right)\left(t-4\right)=t\left(-2\right)$$

Use the distributive property to multiply t+3 by t-4 and combine like terms.

$$t^{2}-t-12=t\left(-2\right)$$

Subtract t\left(-2\right) from both sides.

$$t^{2}-t-12-t\left(-2\right)=0$$

Combine -t and -t\left(-2\right) to get t.

$$t^{2}+t-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$t^{2}+t=12$$

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.

$$t^{2}+t+\left(\frac{1}{2}\right)^{2}=12+\left(\frac{1}{2}\right)^{2}$$

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

$$t^{2}+t+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$t^{2}+t+\frac{1}{4}=\frac{49}{4}$$

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}.

$$\left(t+\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$t+\frac{1}{2}=\frac{7}{2} t+\frac{1}{2}=-\frac{7}{2}$$

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

$$t=3 t=-4$$

Variable t cannot be equal to 3.

$$t=-4$$