(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 $2-3i$.
Best Answer
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.