Solve 10/t^2-25-1/10=1/t-5

$$\frac { 10 } { t ^ { 2 } - 25 } - \frac { 1 } { 10 } = \frac { 1 } { t - 5 }$$

Solve for t

$$t=-15$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by 10\left(t-5\right)\left(t+5\right), the least common multiple of t^{2}-25,10,t-5.

$$10\times 10+10\left(t-5\right)\left(t+5\right)\left(-\frac{1}{10}\right)=10t+50$$

Multiply 10 and 10 to get 100.

$$100+10\left(t-5\right)\left(t+5\right)\left(-\frac{1}{10}\right)=10t+50$$

Multiply 10 and -\frac{1}{10} to get -1.

$$100-\left(t-5\right)\left(t+5\right)=10t+50$$

Use the distributive property to multiply -1 by t-5.

$$100+\left(-t+5\right)\left(t+5\right)=10t+50$$

Use the distributive property to multiply -t+5 by t+5 and combine like terms.

$$100-t^{2}+25=10t+50$$

Add 100 and 25 to get 125.

$$125-t^{2}=10t+50$$

Subtract 10t from both sides.

$$125-t^{2}-10t=50$$

Subtract 50 from both sides.

$$125-t^{2}-10t-50=0$$

Subtract 50 from 125 to get 75.

$$75-t^{2}-10t=0$$

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

$$-t^{2}-10t+75=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt+75. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-75=-75$$

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. List all such integer pairs that give product -75.

$$1,-75 3,-25 5,-15$$

Calculate the sum for each pair.

$$1-75=-74 3-25=-22 5-15=-10$$

The solution is the pair that gives sum -10.

$$a=5 b=-15$$

Rewrite -t^{2}-10t+75 as \left(-t^{2}+5t\right)+\left(-15t+75\right).

$$\left(-t^{2}+5t\right)+\left(-15t+75\right)$$

Factor out t in the first and 15 in the second group.

$$t\left(-t+5\right)+15\left(-t+5\right)$$

Factor out common term -t+5 by using distributive property.

$$\left(-t+5\right)\left(t+15\right)$$

To find equation solutions, solve -t+5=0 and t+15=0.

$$t=5 t=-15$$

Variable t cannot be equal to 5.

$$t=-15$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by 10\left(t-5\right)\left(t+5\right), the least common multiple of t^{2}-25,10,t-5.

$$10\times 10+10\left(t-5\right)\left(t+5\right)\left(-\frac{1}{10}\right)=10t+50$$

Multiply 10 and 10 to get 100.

$$100+10\left(t-5\right)\left(t+5\right)\left(-\frac{1}{10}\right)=10t+50$$

Multiply 10 and -\frac{1}{10} to get -1.

$$100-\left(t-5\right)\left(t+5\right)=10t+50$$

Use the distributive property to multiply -1 by t-5.

$$100+\left(-t+5\right)\left(t+5\right)=10t+50$$

Use the distributive property to multiply -t+5 by t+5 and combine like terms.

$$100-t^{2}+25=10t+50$$

Add 100 and 25 to get 125.

$$125-t^{2}=10t+50$$

Subtract 10t from both sides.

$$125-t^{2}-10t=50$$

Subtract 50 from both sides.

$$125-t^{2}-10t-50=0$$

Subtract 50 from 125 to get 75.

$$75-t^{2}-10t=0$$

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.

$$-t^{2}-10t+75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and 75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\times 75}}{2\left(-1\right)}$$

Square -10.

$$t=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\times 75}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-\left(-10\right)±\sqrt{100+4\times 75}}{2\left(-1\right)}$$

Multiply 4 times 75.

$$t=\frac{-\left(-10\right)±\sqrt{100+300}}{2\left(-1\right)}$$

Add 100 to 300.

$$t=\frac{-\left(-10\right)±\sqrt{400}}{2\left(-1\right)}$$

Take the square root of 400.

$$t=\frac{-\left(-10\right)±20}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$t=\frac{10±20}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{10±20}{-2}$$

Now solve the equation t=\frac{10±20}{-2} when ± is plus. Add 10 to 20.

$$t=\frac{30}{-2}$$

Divide 30 by -2.

$$t=-15$$

Now solve the equation t=\frac{10±20}{-2} when ± is minus. Subtract 20 from 10.

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

Divide -10 by -2.

$$t=5$$

The equation is now solved.

$$t=-15 t=5$$

Variable t cannot be equal to 5.

$$t=-15$$

Steps for Completing the Square

Variable t cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by 10\left(t-5\right)\left(t+5\right), the least common multiple of t^{2}-25,10,t-5.

$$10\times 10+10\left(t-5\right)\left(t+5\right)\left(-\frac{1}{10}\right)=10t+50$$

Multiply 10 and 10 to get 100.

$$100+10\left(t-5\right)\left(t+5\right)\left(-\frac{1}{10}\right)=10t+50$$

Multiply 10 and -\frac{1}{10} to get -1.

$$100-\left(t-5\right)\left(t+5\right)=10t+50$$

Use the distributive property to multiply -1 by t-5.

$$100+\left(-t+5\right)\left(t+5\right)=10t+50$$

Use the distributive property to multiply -t+5 by t+5 and combine like terms.

$$100-t^{2}+25=10t+50$$

Add 100 and 25 to get 125.

$$125-t^{2}=10t+50$$

Subtract 10t from both sides.

$$125-t^{2}-10t=50$$

Subtract 125 from both sides.

$$-t^{2}-10t=50-125$$

Subtract 125 from 50 to get -75.

$$-t^{2}-10t=-75$$

Divide both sides by -1.

$$\frac{-t^{2}-10t}{-1}=-\frac{75}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\left(-\frac{10}{-1}\right)t=-\frac{75}{-1}$$

Divide -10 by -1.

$$t^{2}+10t=-\frac{75}{-1}$$

Divide -75 by -1.

$$t^{2}+10t=75$$

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.

$$t^{2}+10t+5^{2}=75+5^{2}$$

Square 5.

$$t^{2}+10t+25=75+25$$

Add 75 to 25.

$$t^{2}+10t+25=100$$

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}.

$$\left(t+5\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+5\right)^{2}}=\sqrt{100}$$

Simplify.

$$t+5=10 t+5=-10$$

Subtract 5 from both sides of the equation.

$$t=5 t=-15$$

Variable t cannot be equal to 5.

$$t=-15$$