Question

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

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

Short answer

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

Step-by-step solution

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 x2−3x−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.