Math question please ? Complex numbers?

Question

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?

Answer 1

Why can't discriminant be negative?

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

Answer 2

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!

Answer 3

"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!!!

Answer 4

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, $\sqrt{-4} = 2i$