Solve t/t-2+1/t+2=8/t^2-4

$$\frac { t } { t - 2 } + \frac { 1 } { t + 2 } = \frac { 8 } { t ^ { 2 } - 4 }$$

Solve for t

$$t=-5$$

Steps Using Factoring

Variable t cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(t-2\right)\left(t+2\right), the least common multiple of t-2,t+2,t^{2}-4.

$$\left(t+2\right)t+t-2=8$$

Use the distributive property to multiply t+2 by t.

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

Combine 2t and t to get 3t.

$$t^{2}+3t-2=8$$

Subtract 8 from both sides.

$$t^{2}+3t-2-8=0$$

Subtract 8 from -2 to get -10.

$$t^{2}+3t-10=0$$

To solve the equation, factor t^{2}+3t-10 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.

$$a+b=3 ab=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

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

$$\left(t-2\right)\left(t+5\right)$$

To find equation solutions, solve t-2=0 and t+5=0.

$$t=2 t=-5$$

Variable t cannot be equal to 2.

$$t=-5$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(t-2\right)\left(t+2\right), the least common multiple of t-2,t+2,t^{2}-4.

$$\left(t+2\right)t+t-2=8$$

Use the distributive property to multiply t+2 by t.

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

Combine 2t and t to get 3t.

$$t^{2}+3t-2=8$$

Subtract 8 from both sides.

$$t^{2}+3t-2-8=0$$

Subtract 8 from -2 to get -10.

$$t^{2}+3t-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-10\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite t^{2}+3t-10 as \left(t^{2}-2t\right)+\left(5t-10\right).

$$\left(t^{2}-2t\right)+\left(5t-10\right)$$

Factor out t in the first and 5 in the second group.

$$t\left(t-2\right)+5\left(t-2\right)$$

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

$$\left(t-2\right)\left(t+5\right)$$

To find equation solutions, solve t-2=0 and t+5=0.

$$t=2 t=-5$$

Variable t cannot be equal to 2.

$$t=-5$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(t-2\right)\left(t+2\right), the least common multiple of t-2,t+2,t^{2}-4.

$$\left(t+2\right)t+t-2=8$$

Use the distributive property to multiply t+2 by t.

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

Combine 2t and t to get 3t.

$$t^{2}+3t-2=8$$

Subtract 8 from both sides.

$$t^{2}+3t-2-8=0$$

Subtract 8 from -2 to get -10.

$$t^{2}+3t-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-3±\sqrt{3^{2}-4\left(-10\right)}}{2}$$

Square 3.

$$t=\frac{-3±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$t=\frac{-3±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$t=\frac{-3±\sqrt{49}}{2}$$

Take the square root of 49.

$$t=\frac{-3±7}{2}$$

Now solve the equation t=\frac{-3±7}{2} when ± is plus. Add -3 to 7.

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

Divide 4 by 2.

$$t=2$$

Now solve the equation t=\frac{-3±7}{2} when ± is minus. Subtract 7 from -3.

$$t=-\frac{10}{2}$$

Divide -10 by 2.

$$t=-5$$

The equation is now solved.

$$t=2 t=-5$$

Variable t cannot be equal to 2.

$$t=-5$$

Steps for Completing the Square

Variable t cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(t-2\right)\left(t+2\right), the least common multiple of t-2,t+2,t^{2}-4.

$$\left(t+2\right)t+t-2=8$$

Use the distributive property to multiply t+2 by t.

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

Combine 2t and t to get 3t.

$$t^{2}+3t-2=8$$

Add 2 to both sides.

$$t^{2}+3t=8+2$$

Add 8 and 2 to get 10.

$$t^{2}+3t=10$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+3t+\left(\frac{3}{2}\right)^{2}=10+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+3t+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$t^{2}+3t+\frac{9}{4}=\frac{49}{4}$$

Factor t^{2}+3t+\frac{9}{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}.

$$\left(t+\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$t+\frac{3}{2}=\frac{7}{2} t+\frac{3}{2}=-\frac{7}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$t=2 t=-5$$

Variable t cannot be equal to 2.

$$t=-5$$