Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to solve this equation $5z^2+6z+2=0$ where $z$ is a complex number.. I tried writing $z=\alpha+\beta i$ but still nothing..I tried finding the roots but the discriminant is negative $= 36-4\cdot 5 \cdot 2 =-4$ what do I do?

share|cite|improve this question
The discriminant is negative --- yes, that's the first idea of complex numbers, they let you take the square root of a negative number! – Gerry Myerson Dec 9 '12 at 11:21

Why can't discriminant be negative?

$$z=\frac{-6\pm\sqrt{6^2-4\cdot 5\cdot 2}}{2\cdot 5}=\frac{-6\pm2i}{10}=\frac{-3\pm i}5$$

share|cite|improve this answer
you need a $\pm$ instead of a $+$ in your second-to-last step. I can't edit such a small typo, hence the comment. – rubenvb Dec 9 '12 at 11:22
@rubenvb, sorryfor the typo,rectified. – lab bhattacharjee Dec 9 '12 at 11:25

First, consider to use / learn latex notation to have a more clear representation of your problem.

You have $5z^2 + 6z + 2 = 0$, then you simply apply the formula:

$$ z_{1,2} = \frac{-6 \pm \sqrt{36 - 40}}{10} = \frac{-6 \pm 2i}{10} = \frac{-3 \pm i}{5}$$

so you obtain that $z_1 = \frac{-3-i}{5}$ and $z_2 = \frac{-3+i}{5}$.

Remember that you are working in the complex field, so you don't have any problem with a negative discriminant! In a certain way, complex number were born to deal with those situations. In fact, you get that $\sqrt{36-40} = \sqrt{-4} = \sqrt{-1}\sqrt{4} = 2i$ and $i$ is a "member" of the complex numbers!

share|cite|improve this answer

"I tried writing $z=α+βi$ but still nothing ..."

Out of curiousity, I put $z=\alpha+\beta i\ (\beta\not=0)$ and found that ... \begin{align} &5(\alpha^2-\beta^2+2\alpha\beta i)+6(\alpha+\beta i)+2=0,\\ \Leftrightarrow&5(\alpha^2-\beta^2) + 6\alpha + 2 + (10\alpha\beta+6\beta)i=0,\\ \Leftrightarrow&\begin{cases}(a): 5(\alpha^2-\beta^2) + 6\alpha + 2 = 0,\\ (b): (10\alpha + 6)\beta=0.\end{cases} \end{align} As $\beta\neq0$, (b) gives $\alpha = -3/5$. Hence from (a) we get $\beta^2 = \frac{1}{25}$, or $\beta=\pm1/5$. Surprise!!!

share|cite|improve this answer

This type of equations can be solved in the same way as second degree equations with real roots, just remembering $\sqrt{-a^2} = a i$ when a is any positive real number.

In your case, $\sqrt{-4} = 2i$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.