Chapter 12, Problem 71CLR

To determine

To solve: The given equation, inequality, or system, assuming that no denominator is zero.

Answer to Problem 71CLR

The solution is $\left\{\frac{5+\sqrt{13}}{6},\frac{5-\sqrt{13}}{6}\right\}$ .

Explanation of Solution

Given:

The equation is $3t^2-5t+1=0$ .

Calculation:

For an equation $a{x}^2+bx+c=0$ , the solutions can be determined using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Compare the given equation $3t^2-5t+1=0$ with the standard form $a{x}^2+bx+c=0$ to determine the values of a , b and c .

  $\begin{array}{c}a=3\\ b=-5\\ c=1\end{array}$

Substitute the values of a, b and c in the quadratic formula and solve as shown below:

  $\begin{array}{c}x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\left(3\right)\left(1\right)}}{2\left(3\right)}\\ =\frac{5\pm \sqrt{25-12}}{6}\\ =\frac{5\pm \sqrt{13}}{6}\\ =\frac{5+\sqrt{13}}{6},\frac{5-\sqrt{13}}{6}\end{array}$

So, the solution of the given equation is $\left\{\frac{5+\sqrt{13}}{6},\frac{5-\sqrt{13}}{6}\right\}$ .