Chapter 9, Problem 22MCQ

To determine

Solve the equation by using Quadratic formula.

Answer to Problem 22MCQ

Solutions are $-3\text{and}6$

Explanation of Solution

Given:

The equation: $x^2-3x-18=0$

Concept Used:

Quadratic equation: $a{x}^2+bx+c=0,\text{then}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The equation: $x^2-3x-18=0$

  $\begin{array}{l}a{x}^2+bx+c=0,\text{then}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x^2-3x-18=0[\text{make}\text{right}\text{side}\text{equal}\text{to}\text{0}]\\ a=1;b=-3;c=-18\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(-3)\pm \sqrt{{(-3)}^2-4·1·-18}}{2·1}=\frac{3\pm \sqrt{9+72}}{2}=\frac{3\pm \sqrt{81}}{2}=\frac{3\pm 9}{2}\\ x=\frac{3-9}{2}\text{and}\frac{3+9}{2}\\ x=-3\text{and}6\end{array}$

Solutions are $-3\text{and}6$

Thus, solutions are $-3\text{and}6$