# How to find the imaginary roots of polynomials

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.

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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 $x=\cfrac{1 + i\sqrt{15}}{4}$ $OR$ $\cfrac{1- i\sqrt{15}}{4}$

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Solving x1: $\dfrac{(1+\sqrt(1^2-4*2*2))}{4} 4$) <=> $1/4-sqr(15)/4i$? So when sqr(-15) appears I have to think, how can i solve this with complex numbers? –  Alek Oliver Oct 30 '12 at 17:54
@AlekOliver can you edit your comment? –  user31280 Oct 30 '12 at 17:55
Still working on it, not that used to Latex yet :) –  Alek Oliver Oct 30 '12 at 17:56
@AlekOliver $\sqrt{-15} = i\sqrt{15}$ and your result should be $\cfrac{1 \pm i\sqrt{15}}{4}$ –  user31280 Oct 30 '12 at 17:57
Solving x1= $\dfrac{(1+\sqrt(1^2-4*2*2))}{4}$) = $\dfrac{(1/4-sqrt(15)}{4}$ = $\dfrac{(1/4-isqrt(15)}{4}$ $\\$ x2 = $\dfrac{(1/4+isqrt(15)}{4}$ Like this? F0la Yinka thanks:) –  Alek Oliver Oct 30 '12 at 18:00

$$2x^2 - x +2 = 2\left(x^2 - \dfrac{x}2 + 1 \right) = 2\left(\underbrace{x^2 - 2 \times x \times \dfrac14 + \left(\dfrac14 \right)^2}_{a^2 - 2ab + b^2 = (a-b)^2} - \left(\dfrac14 \right)^2 + 1 \right) = 2 \left( \left(x- \dfrac14 \right)^2 + \dfrac{15}{16}\right)$$ 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, $$x = \dfrac{1 \pm i \sqrt{15}}4$$

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Thanks for the alternative solutions baka-sorouh & @marvis –  Alek Oliver Oct 30 '12 at 18:17
If you spot it you can complete the square as $\frac 1 8 (16x^2-8x+16) = \frac 1 8 \left( (4x-1)^2+15\right)$. The factor is $4a$. –  Mark Bennet Oct 30 '12 at 18:17
@AlekOliver: The way Marvis suggested is full of practical methods you can use for other second order polynomial equation. Try to reflect it.+1 –  Babak S. Oct 30 '12 at 18:28
@BabakSorouh: I'm simply sticking with Fola's solution cause I'm studying for a multiply choice test. However I will try to learn the other way later this week, when I get some spare time, as I can see the benefits of it. –  Alek Oliver Oct 31 '12 at 6:33

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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+1 Nice and original idea. –  DonAntonio Nov 29 '12 at 13:33
@DonAntonio: You are always welcome to me, Don. –  Babak S. Nov 29 '12 at 14:10