Chapter 9.6, Problem 21PPS

To determine

The solutions of the equation $5x^2+21x=-18$ using quadratic formula

Answer to Problem 21PPS

  $x=\frac{-6}{5},\text{ and }x=-3$

Explanation of Solution

Given:

The equation,

  $5x^2+21x=-18$

Concept Used:

Quadratic formula: The roots of a quadratic equation $a{x}^2+bx+c=0;\text{ where }a\ne 0$ are:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

In order to solve the given equation $5x^2+21x+18=0$ , since it is in the descending order of exponents, so comparing with the general equation $a{x}^2+bx+c=0$ , the value of the coefficient a, b, and c is given by

  $a=5,b=21,\text{ and }c=18$

Now, above quadratic equation $5y^2+21y+18=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(21\right)\pm \sqrt{{\left(21\right)}^2-4\left(5\right)\left(18\right)}}{2\left(5\right)}\\ =\frac{-21\pm \sqrt{441-360}}{10}\\ =\frac{-21\pm \sqrt{81}}{10}\\ =\frac{-21\pm 9}{10}\\ x=\frac{-21}{10}+\frac{9}{10},\text{ and }x=\frac{-21}{10}-\frac{9}{10}\\ \Rightarrow x=\frac{-12}{10},\text{ and }x=\frac{-30}{10}\\ \Rightarrow x=\frac{-6}{5},\text{ and }x=-3\end{array}$

Thus, the solutions of the given question are, $x=\frac{-6}{5},\text{ and }x=-3$ .