{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
{{ searchError }}
search
{{ courseTrack.displayTitle }}
{{ printedBook.courseTrack.name }} {{ printedBook.name }}

# Using the Quadratic Formula to find Complex Roots

## Using the Quadratic Formula to find Complex Roots 1.7 - Solution

We'll use the Quadratic Formula to solve the given quadratic equation. $\begin{gathered} {\color{#0000FF}{a}}x^2+{\color{#FF0000}{b}}x+{\color{#009600}{c}}=0 \quad \Leftrightarrow \quad x=\dfrac{\text{-} {\color{#FF0000}{b}}\pm \sqrt{{\color{#FF0000}{b}}^2-4{\color{#0000FF}{a}}{\color{#009600}{c}}}}{2{\color{#0000FF}{a}}} \end{gathered}$ We first need to identify the values of ${\color{#0000FF}{a}},$ ${\color{#FF0000}{b}},$ and ${\color{#009600}{c}}.$ $\begin{gathered} {\color{#0000FF}{2}}x^2+({\color{#FF0000}{\text{-} 16}})x+{\color{#009600}{50}}=0 \end{gathered}$ We see that ${\color{#0000FF}{a}}={\color{#0000FF}{2}},$ ${\color{#FF0000}{b}}={\color{#FF0000}{\text{-} 16}},$ and ${\color{#009600}{c}}={\color{#009600}{50}}.$ Let's substitute these values into the Quadratic Formula.
$x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{\text{-} ({\color{#FF0000}{\text{-} 16}})\pm\sqrt{({\color{#FF0000}{\text{-} 16}})^2-4({\color{#0000FF}{2}})({\color{#009600}{50}})}}{2({\color{#0000FF}{2}})}$
Solve for $x$ and Simplify
$x=\dfrac{16\pm\sqrt{(\text{-} 16)^2-4(2)(50)}}{2(2)}$
$x=\dfrac{16\pm\sqrt{256-4(2)(50)}}{2(2)}$
$x=\dfrac{16\pm\sqrt{256-4(100)}}{4}$
$x=\dfrac{16\pm\sqrt{256-400}}{4}$
$x=\dfrac{16\pm\sqrt{\text{-} 144}}{4}$
$x=\dfrac{16\pm\sqrt{\text{-} 1 \cdot 144}}{4}$
$x=\dfrac{16 \pm \sqrt{\text{-} 1} \cdot \sqrt{144}}{4}$
$x=\dfrac{16 \pm i \cdot \sqrt{144}}{4}$
$x=\dfrac{16 \pm i \cdot 12}{4}$
$x=\dfrac{4(4 \pm i \cdot 3)}{4}$
$x=4 \pm i \cdot3$
$x=4 \pm 3i$
Using the Quadratic Formula, we found that the solutions of the given equation are $x=4 \pm 3i.$ Thus, the solutions are $x_1=4 + 3i$ and $x_2=4- 3i.$