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

## Problem

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}

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 ?

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

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 $$3a+1=0.$$

Edit

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.

• Could you explain why ?
– Tuki
Oct 10, 2018 at 14:04
• 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).
– 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

$$2p=-(3a+1)$$

$$p^2+q^2=81$$

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. $$3a+1=0$$