Q22. - Algebra 1

Question

Solve the equation by completing the square. Round to the nearest tenth.

$x^2-3x-18=0$

Step-by-Step Solution

Answer

$\mathit{x}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{6}$ are the solutions of the given equation.

Step 1. Define the standard form of the quadratic equation.

A quadratic equation, which is written in the form, $a{x}^2+bx+c=0$, where, $a\ne 0$ is called the standard form of the quadratic equation.

Step 2. Define the quadratic formula.

For the quadratic equation $a{x}^2+bx+c=0,$ where, $a\ne 0$ is given by 

 $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Step 3. Solve the equation x 2 − 3 x − 18 = 0 by completing the square.

Compare the quadratic equation $x^2-3x-18=0$ with the quadratic equation $a{x}^2+bx+c=0$.

$a=1,b=-3,c=-18$

Substitute, $a=1,b=-3$ and $c=-18$ in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$$\begin{gathered} x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\left(1\right)\left(-18\right)}}{2\left(1\right)} \\ x=\frac{3\pm \sqrt{9+72}}{2} \\ x=\frac{3\pm \sqrt{81}}{2} \\ x=\frac{3\pm 9}{2} \\ x=\frac{3-9}{2},\frac{3+9}{2} \\ x=\frac{-6}{2},\frac{12}{2} \\ x=-3,6 \end{gathered}$$

Hence, $x=-3,6$ are the solutions of the given equation.