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Using the Quadratic Formula to find Complex Roots

Using the Quadratic Formula to find Complex Roots 1.7 - Solution

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We'll use the Quadratic Formula to solve the given quadratic equation. ax2+bx+c=0x=-b±b24ac2a\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 a,{\color{#0000FF}{a}}, b,{\color{#FF0000}{b}}, and c.{\color{#009600}{c}}. 2x2+(-16)x+50=0\begin{gathered} {\color{#0000FF}{2}}x^2+({\color{#FF0000}{\text{-} 16}})x+{\color{#009600}{50}}=0 \end{gathered} We see that a=2,{\color{#0000FF}{a}}={\color{#0000FF}{2}}, b=-16,{\color{#FF0000}{b}}={\color{#FF0000}{\text{-} 16}}, and c=50.{\color{#009600}{c}}={\color{#009600}{50}}. Let's substitute these values into the Quadratic Formula.
x=-b±b24ac2ax=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a}
x=-(-16)±(-16)24(2)(50)2(2)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 xx and Simplify
x=16±(-16)24(2)(50)2(2)x=\dfrac{16\pm\sqrt{(\text{-} 16)^2-4(2)(50)}}{2(2)}
x=16±2564(2)(50)2(2)x=\dfrac{16\pm\sqrt{256-4(2)(50)}}{2(2)}
x=16±2564(100)4x=\dfrac{16\pm\sqrt{256-4(100)}}{4}
x=16±2564004x=\dfrac{16\pm\sqrt{256-400}}{4}
x=16±-1444x=\dfrac{16\pm\sqrt{\text{-} 144}}{4}
x=16±-11444x=\dfrac{16\pm\sqrt{\text{-} 1 \cdot 144}}{4}
x=16±-11444x=\dfrac{16 \pm \sqrt{\text{-} 1} \cdot \sqrt{144}}{4}
x=16±i1444x=\dfrac{16 \pm i \cdot \sqrt{144}}{4}
x=16±i124x=\dfrac{16 \pm i \cdot 12}{4}
x=4(4±i3)4x=\dfrac{4(4 \pm i \cdot 3)}{4}
x=4±i3x=4 \pm i \cdot3
x=4±3ix=4 \pm 3i
Using the Quadratic Formula, we found that the solutions of the given equation are x=4±3i.x=4 \pm 3i. Thus, the solutions are x1=4+3ix_1=4 + 3i and x2=43i.x_2=4- 3i.