Solve 18/t^2+1/t=4

$$\frac { 18 } { t ^ { 2 } } + \frac { 1 } { t } = 4$$

Solve for t

$$t = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

$$t=-2$$

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^{2}, the least common multiple of t^{2},t.

$$18+t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$18+t-4t^{2}=0$$

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

$$-4t^{2}+t+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4t^{2}+at+bt+18. To find a and b, set up a system to be solved.

$$a+b=1 ab=-4\times 18=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$a=9 b=-8$$

Rewrite -4t^{2}+t+18 as \left(-4t^{2}+9t\right)+\left(-8t+18\right).

$$\left(-4t^{2}+9t\right)+\left(-8t+18\right)$$

Factor out -t in the first and -2 in the second group.

$$-t\left(4t-9\right)-2\left(4t-9\right)$$

Factor out common term 4t-9 by using distributive property.

$$\left(4t-9\right)\left(-t-2\right)$$

To find equation solutions, solve 4t-9=0 and -t-2=0.

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

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^{2}, the least common multiple of t^{2},t.

$$18+t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$18+t-4t^{2}=0$$

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.

$$-4t^{2}+t+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 1 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-4\right)\times 18}}{2\left(-4\right)}$$

Square 1.

$$t=\frac{-1±\sqrt{1-4\left(-4\right)\times 18}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$t=\frac{-1±\sqrt{1+16\times 18}}{2\left(-4\right)}$$

Multiply 16 times 18.

$$t=\frac{-1±\sqrt{1+288}}{2\left(-4\right)}$$

Add 1 to 288.

$$t=\frac{-1±\sqrt{289}}{2\left(-4\right)}$$

Take the square root of 289.

$$t=\frac{-1±17}{2\left(-4\right)}$$

Multiply 2 times -4.

$$t=\frac{-1±17}{-8}$$

Now solve the equation t=\frac{-1±17}{-8} when ± is plus. Add -1 to 17.

$$t=\frac{16}{-8}$$

Divide 16 by -8.

$$t=-2$$

Now solve the equation t=\frac{-1±17}{-8} when ± is minus. Subtract 17 from -1.

$$t=-\frac{18}{-8}$$

Reduce the fraction \frac{-18}{-8} to lowest terms by extracting and canceling out 2.

$$t=\frac{9}{4}$$

The equation is now solved.

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

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^{2}, the least common multiple of t^{2},t.

$$18+t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$18+t-4t^{2}=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$t-4t^{2}=-18$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4t^{2}+t=-18$$

Divide both sides by -4.

$$\frac{-4t^{2}+t}{-4}=-\frac{18}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$t^{2}+\frac{1}{-4}t=-\frac{18}{-4}$$

Divide 1 by -4.

$$t^{2}-\frac{1}{4}t=-\frac{18}{-4}$$

Reduce the fraction \frac{-18}{-4} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{1}{4}t+\left(-\frac{1}{8}\right)^{2}=\frac{9}{2}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{1}{4}t+\frac{1}{64}=\frac{9}{2}+\frac{1}{64}$$

Add \frac{9}{2} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{1}{4}t+\frac{1}{64}=\frac{289}{64}$$

Factor t^{2}-\frac{1}{4}t+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{8}\right)^{2}=\frac{289}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{8}\right)^{2}}=\sqrt{\frac{289}{64}}$$

Simplify.

$$t-\frac{1}{8}=\frac{17}{8} t-\frac{1}{8}=-\frac{17}{8}$$

Add \frac{1}{8} to both sides of the equation.

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