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Using the Quadratic Formula to find Complex Roots

Using the Quadratic Formula to find Complex Roots 1.1 - Solution

a
In the Quadratic Formula, ${\color{#FF0000}{b^2-4ac}}$ is the discriminant. $\begin{gathered} ax^2+bx+c=0 \quad \Leftrightarrow \quad x=\dfrac{\text{-} b\pm\sqrt{{\color{#FF0000}{b^2-4ac}}}}{2a} \end{gathered}$ Let's first rewrite the given equation in standard form.
$0.8x^2+2.6x=\text{-} 3.2$
$0.8x^2+2.6x+3.2=0$
Having rewritten the equation, we can identify the values of ${\color{#0000FF}{a}},$ ${\color{#FF0000}{b}},$ and ${\color{#009600}{c}}.$ $\begin{gathered} 0.8x^2+2.6x=\text{-} 3.2 \ \ \Leftrightarrow \ \ {\color{#0000FF}{0.8}}x^2+{\color{#FF0000}{2.6}}x+{\color{#009600}{3.2}}=0 \end{gathered}$ Now, let's evaluate the discriminant.
$b^2-4ac$
${\color{#FF0000}{2.6}}^2-4({\color{#0000FF}{0.8}})({\color{#009600}{3.2}})$
Simplify
$6.76-4(0.8)(3.2)$
$6.76-3.2(3.2)$
$6.76-10.24$
$\text{-} 3.48$
The discriminant is $\text{-} 3.48.$
b

We want to use the discriminant of the given quadratic equation to determine the number and type of the roots. If we don't want to know the exact values of the roots, we only need to work with the discriminant. From Part A, we know that the discriminant of the given equation is $\text{-} 3.48.$ \begin{aligned} \textbf{Equation: }& 0.8x^2+2.6x=\text{-} 3.2\\ \textbf{Discriminant: }& \text{-} 3.48 \end{aligned} Since the discriminant is less than zero, the quadratic equation has two complex roots.

c
We will use the Quadratic Formula to find the exact solutions of the given equation. $\begin{gathered} x=\dfrac{\text{-}{\color{#FF0000}{b}}\pm\sqrt{{\color{#FF0000}{b}}^2-4{\color{#0000FF}{a}}{\color{#009600}{c}}}}{2{\color{#0000FF}{a}}} \end{gathered}$ Recall that we have already identified the values of ${\color{#0000FF}{a}},$ ${\color{#FF0000}{b}},$ and ${\color{#009600}{c}}$ in Part A, as well as the discriminant, $b^2-4ac.$ $\begin{gathered} {\color{#0000FF}{a}}={\color{#0000FF}{0.8}},\,{\color{#FF0000}{b}}={\color{#FF0000}{2.6}},\,{\color{#009600}{c}}={\color{#009600}{3.2}}\\ \textbf{Discriminant: } \textcolor{magenta}{\text{-} 3.48} \end{gathered}$ Let's substitute these values into the Quadratic Formula.
$x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{\text{-}{\color{#FF0000}{2.6}}\pm\sqrt{\textcolor{magenta}{\text{-} 3.48}}}{2({\color{#0000FF}{0.8}})}$
Simplify
$x=\dfrac{\text{-} 2.6\pm\sqrt{\text{-} 1\cdot 4\cdot 0.87}}{2(0.8)}$
$x=\dfrac{\text{-} 2.6\pm\sqrt{\text{-} 1}\cdot\sqrt{4}\cdot\sqrt{0.87}}{2(0.8)}$
$x=\dfrac{\text{-} 2.6\pm i\sqrt{4}\cdot\sqrt{0.87}}{2(0.8)}$
$x=\dfrac{\text{-} 2.6\pm 2i\sqrt{0.87}}{2(0.8)}$
$x=\dfrac{2(\text{-} 1.3\pm i\sqrt{0.87})}{2(0.8)}$
$x=\dfrac{\text{-} 1.3\pm i\sqrt{0.87}}{0.8}$
Using the Quadratic Formula, we found that the solutions of the given equation are $x=\frac{\text{-} 1.3\pm i\sqrt{0.87}}{0.8}.$