(Biquadratic $\rightarrow$ Quartic (degree 4))

The Question: (from a book i am practicing from)

Find the nature of the roots of the equation $$f(x) = 45 x^4-144 x^3+146 x^2-56 x+12=0$$

(By nature i mean...Real/Imaginary)

My attempt:

Well my first attempt was knowing at least one root by Hit-and-Trial. So that it is of the form $$f(x)=(x-\alpha)(ax^3+bx^2+cx+d)=0$$

And then I would have solved the cubic the same way and finally having $$(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)=0$$

But Hit-and-Trial doesn't seem to work here.

Next attempt:

I tried raw graphing of the function by using maxima/minima.

I differentiated $f(x)$ to get $$f'(x)=45x^3-108x^2+73x-14$$

I did $f'(x)=0$ and Thought I would arrive at some values of $x$ resulting in a graph like :


But I couldn't as I could not even solve this cubic equation!

So how should I solve this problem? Please also help me about drawing the graph of a biquadratic equation. Thanks

P.S. - Here is a link to wolfram alpha for this equation : http://goo.gl/zs4zDa [It also has the roots as $1.1718,1.5671,0.23-0.3i,0.23+0.3i$]

EDIT - Please do not think that I will be able to guess $1.1718$ while solving the quadratic or $\frac{1}{3}$ while solving its derivative (the cubic)! Also I want to ask this question for a general biquadratic equation where all roots may not be real. So no Easy Guesses!! In some cases the roots may be like $\frac{1+\sqrt{3}}{2}$ or something like $2-3i$.

  • 1
    $\begingroup$ You're not meant to find the roots explicitly. All you need to do is show that there is at least one real and one complex root. Then you know there is another complex conjugate and another real root. This gives a total of 2 complex and 2 real roots. $\endgroup$
    – Zain Patel
    Jul 8 '15 at 15:30
  • $\begingroup$ @ZainPatel Ya i know...that's why i was also trying to do it graphically...BTW thanks for the 'complex conjugate' idea! I had forgotten it $\endgroup$
    – NeilRoy
    Jul 8 '15 at 15:32
  • 1
    $\begingroup$ There is no biquadratic equation here, as there are cubic and linear terms. $\endgroup$
    – user65203
    Jul 8 '15 at 15:42
  • $\begingroup$ The general method for solving such equations is quite complicated, you don't want to go there. $\endgroup$
    – Axel
    Jul 16 '15 at 8:48
  • 1
    $\begingroup$ Well I think @ZainPatel 's answer is valid. $\endgroup$
    – Axel
    Jul 16 '15 at 13:48

The derivative of your quartic is $$4(45x^3 - 108x^2 + 73x - 14) = 4(3x-2)(3x-1)(5x-7)$$

So the stationary points are at $x = \frac{1}{3}, \frac{2}{3}, \frac{7}{5}$, does this help you draw the graph?

In addition to this, we have the second derivative as $$540x^2-864x + 292$$ that will allow you to check whether your stationary points are minimums or maximums.

enter image description here

  • $\begingroup$ how did you factorize the cubic eq.? Also the whole point of this question is for how to deal with a situation where it is NON-factorizable. The eq i gave is just an example... $\endgroup$
    – NeilRoy
    Jul 8 '15 at 15:38
  • $\begingroup$ The quartic is non-factorizable, but it's derivative is factorisable. You do hit-and-trial to guess one root and then divide the cubic by $x- \frac{1}{3}$ to get a quadratic that you can solve using the quad. formula for the other two stationary points. $\endgroup$
    – Zain Patel
    Jul 8 '15 at 15:41
  • $\begingroup$ How am i supposed to guess $\frac{1}{3}$ ? Or for example if the root is $\frac{21}{22}$...What am i gonna do then? $\endgroup$
    – NeilRoy
    Jul 8 '15 at 15:53
  • 7
    $\begingroup$ Have you heard of the rational roots theorem? $\endgroup$
    – Zain Patel
    Jul 8 '15 at 15:54
  • 3
    $\begingroup$ In that case, you'll have to resort to numerical methods like interval bisection. $\endgroup$
    – Zain Patel
    Jul 8 '15 at 15:57

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