I need help with a competition math problem please.

Question

I am a competition math student, and I was doing a practice test when I found this problem: "The fourth degree polynomial equation $x^4 - 7x^3 + 4x^2 + 7x - 4 = 0$ has four real roots a, b, c and d. What is the value of the sum $1/a + 1/b + 1/c + 1/d$? Express your answer as a common fraction." First of all, I don't think this polynomial can be factored, so I don't know how to find the reciprocals of the roots. I need help and any help would be appreciated! Note: When they say "Express your answer as a common fraction.", it means that the answer IS a fraction. I tried asking this question already, but I worded it differently, and they didn't get a fraction so I knew it was wrong. This is the link to my other question: Link

Answer 1

Suppose we have a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_0$ with roots $r_1,r_2,\dots, r_n$. We claim that the polynomial with roots $\frac{1}{r_1},\frac{1}{r_2},\dots,\frac{1}{r_n}$ is simply $g(x)=a_0x^n+a_1x^{n-1}+\dots + a_n$. Why? Because if you observe that the roots of the polynomial $f(\frac{1}{x})$ are the reciprocals of the roots of $f(x)$, then our desired polynomial is $$\begin{align*}g(x)&=x^nf(1/x)\\&=x^n\left(\frac{a_n}{x^n}+\frac{a_{n-1}} {x^{n-1}}+\cdots+\frac{a_1}{x}+a_0\right)\\&=a_0x^n+a_1x^{n-1}+\cdots+a_n,\end{align*}$$

The desired polynomial can be found by simply 'reversing the coefficients' as shown above. Hence the polynomial with roots $\frac{1}{a},\frac 1b, \frac 1c,$ and $\frac 1d$ is $$-4x^4+7x^3+4x^2-7x+1$$ Applying Vieta's Formulae to this, we find that the sum of the roots is $-\frac{-7}{4}=\frac{7}{4}$.

Answer 2

The trick is to write $$ \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}=\frac{bcd+acd+abd+abc}{abcd} $$ write $f(x)=x^4-7x^3+4x^2+7x-4=(x-a)(x-b)(x-c)(x-d)$ and use Vieta's relations.

For example, by multiplying out we see at once that $abcd=-4$ and $-(bcd+acd+abd+abc)=7.$