I'm looking for a simple way to calculate roots in the complex numbers. I'm having the polynomial $2x^2-x+2$ I know that the result is $1/4-$($\sqrt{15}/4)i$. However I'm not aware of an easy way to get there.
$\endgroup$3 Answers
$\begingroup$$$2x^2 - x +2 = 2\left(x^2 - \dfrac{x}2 + 1 \right) = 2\left(\underbrace{x^2 - 2 \times x \times \dfrac14 + \left(\dfrac14 \right)^2}_{a^2 - 2ab + b^2 = (a-b)^2} - \left(\dfrac14 \right)^2 + 1 \right) = 2 \left( \left(x- \dfrac14 \right)^2 + \dfrac{15}{16}\right)$$ Setting the above to zero, we get that $$\left(x- \dfrac14 \right)^2 + \dfrac{15}{16} = 0 \implies \left(x- \dfrac14 \right)^2 = \dfrac{-15}{16} \implies x - \dfrac14 = \pm\dfrac{i\sqrt{15}}{4}$$ Hence, $$x = \dfrac{1 \pm i \sqrt{15}}4$$
$\endgroup$ 4 $\begingroup$Hint: $x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ for a quadratic equation $ax^2+bx+c=0$ and $i^2=-1$
You have $2x^2-x+2=0$, $a=2$, $b=-1$ and $c=2$, then $$x=\cfrac{-(-1)\pm\sqrt{(-1)^2-4(2)(2)}}{2(2)}=\cfrac{1\pm\sqrt{-15}}{4}=\cfrac{1\pm i\sqrt{15}}{4}$$
and $x=\cfrac{1 + i\sqrt{15}}{4}$ $OR$ $\cfrac{1- i\sqrt{15}}{4}$
$\endgroup$ 5 $\begingroup$You can use the way that @FOla suggested, or you can do as following: $$2x^2-x+2=0\rightarrow2x^2-x+2+\frac{1}{8}-\frac{1}{8}=0\rightarrow(2x^2-x+\frac{1}{8})=\frac{1}{8}\\(\sqrt{2}x-\frac{\sqrt{2}}{4})^2=\frac{1}{8}-2=\frac{-15}{8}\rightarrow\sqrt{2}x-\frac{\sqrt{2}}{4}=\pm\bigg ( \frac{\sqrt{15}I}{2\sqrt{2}}\bigg)$$ Now divide both sides by $\sqrt{2}$. You would see your solutions.
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