Chapter 1.8, Problem 11E

Solution

Solve the quadratic equation $3r^2-8r-9=0$ for $\left(r\right)$ .

The solution for the quadratic equation is $r=\left\{\frac{4+\sqrt{43}}{3},\frac{4-\sqrt{43}}{3}\right\}$ .

Given:

The quadratic equation is $3r^2-8r-9=0$ .

Concept used:

Any quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has the roots $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(1\right)$

Calculation:

The solution $\left(r\right)$ for the quadratic equation $3r^2-8r-9=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(3\right)\left(-9\right)}}{2\left(3\right)}\mathrm{......}\left(\because 3r^2-8r-9=0\right)\\ =\frac{8\pm \sqrt{172}}{6}\\ =\frac{8\pm 2\sqrt{43}}{6}\\ =\left\{\frac{4+\sqrt{43}}{3},\frac{4-\sqrt{43}}{3}\right\}\end{array}$

Hence the solution for the quadratic equation is $r=\left\{\frac{4+\sqrt{43}}{3},\frac{4-\sqrt{43}}{3}\right\}$ .