Solve t+1/t=13/6

$$t + \frac { 1 } { t } = \frac { 13 } { 6 }$$

Solve for t

$$t=\frac{2}{3}\approx 0.666666667$$

$$t = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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 6t, the least common multiple of t,6.

$$6tt+6=13t$$

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

$$6t^{2}+6=13t$$

Subtract 13t from both sides.

$$6t^{2}+6-13t=0$$

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

$$6t^{2}-13t+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6t^{2}+at+bt+6. To find a and b, set up a system to be solved.

$$a+b=-13 ab=6\times 6=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 36.

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

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -13.

$$a=-9 b=-4$$

Rewrite 6t^{2}-13t+6 as \left(6t^{2}-9t\right)+\left(-4t+6\right).

$$\left(6t^{2}-9t\right)+\left(-4t+6\right)$$

Factor out 3t in the first and -2 in the second group.

$$3t\left(2t-3\right)-2\left(2t-3\right)$$

Factor out common term 2t-3 by using distributive property.

$$\left(2t-3\right)\left(3t-2\right)$$

To find equation solutions, solve 2t-3=0 and 3t-2=0.

$$t=\frac{3}{2} t=\frac{2}{3}$$

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 6t, the least common multiple of t,6.

$$6tt+6=13t$$

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

$$6t^{2}+6=13t$$

Subtract 13t from both sides.

$$6t^{2}+6-13t=0$$

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.

$$6t^{2}-13t+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -13 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\times 6}}{2\times 6}$$

Square -13.

$$t=\frac{-\left(-13\right)±\sqrt{169-4\times 6\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$t=\frac{-\left(-13\right)±\sqrt{169-24\times 6}}{2\times 6}$$

Multiply -24 times 6.

$$t=\frac{-\left(-13\right)±\sqrt{169-144}}{2\times 6}$$

Add 169 to -144.

$$t=\frac{-\left(-13\right)±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$t=\frac{-\left(-13\right)±5}{2\times 6}$$

The opposite of -13 is 13.

$$t=\frac{13±5}{2\times 6}$$

Multiply 2 times 6.

$$t=\frac{13±5}{12}$$

Now solve the equation t=\frac{13±5}{12} when ± is plus. Add 13 to 5.

$$t=\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

$$t=\frac{3}{2}$$

Now solve the equation t=\frac{13±5}{12} when ± is minus. Subtract 5 from 13.

$$t=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

$$t=\frac{2}{3}$$

The equation is now solved.

$$t=\frac{3}{2} t=\frac{2}{3}$$

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 6t, the least common multiple of t,6.

$$6tt+6=13t$$

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

$$6t^{2}+6=13t$$

Subtract 13t from both sides.

$$6t^{2}+6-13t=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$6t^{2}-13t=-6$$

Divide both sides by 6.

$$\frac{6t^{2}-13t}{6}=-\frac{6}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$t^{2}-\frac{13}{6}t=-\frac{6}{6}$$

Divide -6 by 6.

$$t^{2}-\frac{13}{6}t=-1$$

Divide -\frac{13}{6}, the coefficient of the x term, by 2 to get -\frac{13}{12}. Then add the square of -\frac{13}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{13}{6}t+\left(-\frac{13}{12}\right)^{2}=-1+\left(-\frac{13}{12}\right)^{2}$$

Square -\frac{13}{12} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{13}{6}t+\frac{169}{144}=-1+\frac{169}{144}$$

Add -1 to \frac{169}{144}.

$$t^{2}-\frac{13}{6}t+\frac{169}{144}=\frac{25}{144}$$

Factor t^{2}-\frac{13}{6}t+\frac{169}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{13}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{13}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$t-\frac{13}{12}=\frac{5}{12} t-\frac{13}{12}=-\frac{5}{12}$$

Add \frac{13}{12} to both sides of the equation.

$$t=\frac{3}{2} t=\frac{2}{3}$$