Define constant $a$ in way that $x^2+(3a+1)x+81=0$ solutions are complex.

After that define $a$ in way that the solutions are strictly imaginary (when real part is $0$)

Attempt to solve

In a way real solutions for this are also complex. Real numbers are subset of complex numbers hence all real solutions are also complex. I think this was intended to interpret as solutions that are form $z=a+ib$ when $(a,b) \in \mathbb{R}, z \in \mathbb{C}$, $b \neq 0$ so solution is complex when it has imaginary component.

$$ x^2+(3a+1)x+81=0 $$

By utilizing quadratic formula we can use discriminant to define for what constant $a$ value equation has only complex solutions.

$$ D=(3a+1)^2-4\cdot 81 $$

if $D<0$ all solutions have to have imaginary part hence they are complex.

Our discriminant is form of a parabola if we would use $a$ as variable.

$$ (3a+1)^2=9a^2+6a+1 $$

if we compute when $D=0$ we can figure out when $D < 0$

$$ 9a^2+6a-323 = 0 $$

$$ a = \frac{ -6 \pm \sqrt{6^2-4 \cdot (-323) \cdot 9} }{ 2 \cdot 9 } $$

$$ a = \frac{ -6 \pm 108 } { 18 } $$

$$ a_1 = \frac{ -19 }{ 3 }, a_2 = \frac{ 17 }{ 3 } $$

So we know that $D < 0$ when:

$$ \frac{-19}{3}<a<\frac{17}{3} $$

Now only problem is i don't know if my solution is valid and how do you define a when solutions have to be strictly imaginary ?

  • 1
    $\begingroup$ By "complex" do you actually mean "not real"? $\endgroup$ Oct 10, 2018 at 14:01
  • $\begingroup$ Yes exactly @AndrésE.Caicedo $\endgroup$
    – Tuki
    Oct 10, 2018 at 14:02

2 Answers 2


Second part

$$x^2+(3a+1)x+81=0\implies x=\frac{-(3a+1)\pm \sqrt{(3a+1)^2-4\cdot 81}}{2}.$$

The roots are strictly imaginary if and only if $3a+1=0.$


The roots are

$$\dfrac{-(3a+1)\pm \sqrt D}{2}$$ where $D$ is a real number. So

$$\sqrt{D}=\pm i\sqrt{|D|}$$ if $D<0.$ Thus

$$\Re \{\dfrac{-(3a+1)\pm \sqrt D}{2}\}=\dfrac{-(3a+1)}{2}. $$

If $D\ge 0$ the solutions are real.

  • $\begingroup$ Could you explain why ? $\endgroup$
    – Tuki
    Oct 10, 2018 at 14:04
  • $\begingroup$ I have added an explanation. With respect to the first part it is correct (the way you proceed, but I didn't check the calculations). $\endgroup$
    – mfl
    Oct 10, 2018 at 14:08

All coefficients are real, so any complex valued roots occur as conjugate pairs, $p\pm i q$.

Considering the sum and product of the roots



If the roots are real $q=0 \Rightarrow p = \pm9 \Rightarrow a = -19/3, 17/3$.

Thus for non-real i.e. complex roots you need $a \neq -19/3$ and $a \neq 17/3$

For strictly imaginary roots, the real part $p =0$ i.e. $3a+1=0$


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