2
$\begingroup$

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:

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

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 ?

$\endgroup$
  • 1
    $\begingroup$ By "complex" do you actually mean "not real"? $\endgroup$ – Andrés E. Caicedo Oct 10 '18 at 14:01
  • $\begingroup$ Yes exactly @AndrésE.Caicedo $\endgroup$ – Tuki Oct 10 '18 at 14:02
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Could you explain why ? $\endgroup$ – Tuki Oct 10 '18 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 '18 at 14:08
0
$\begingroup$

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. $3a+1=0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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