If $\alpha$ and $\beta$ are the zeros of the polynomial $p(s)=3s^2-6s+4$, find the value of

Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $p(s)=3s^2-6s+4$, find the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac{1}{\alpha}+\frac{1}{\beta})+3\alpha\beta$

By calculating, I am finding the answer as $9$, but, the book has the answer as $8$. I did a cross check and my answer seems to be right.

Answer 1

We have $\alpha+\beta=2$ and $\alpha\beta=\frac{4}{3}$.

The first two terms add up to $\frac{\alpha^2+\beta^2}{\alpha\beta}$, which is $\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$. This is $1$.

The term $2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$, that is, $2\left(\frac{\alpha+\beta}{\alpha\beta}\right)$, is equal to $3$, and the term $3\alpha\beta$ is equal to $4$.

The sum of all the terms is $8$.

Answer 2

Rewrite $$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac 1{\alpha}+\frac 1{\beta})+3{\alpha}{\beta}=\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2\frac{\alpha+\beta}{\alpha\beta}+3{\alpha\beta}=\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2\frac SP+3P$$ where $S$ and $P$ are the sum and the product of the roots.

Now $$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{S^2-2P}P$$ With $S=2$ and $P=\frac 43$. Then, the result is $8$.

Edit

Another possible way : since the equation has no real root, let us set $\alpha=a+i b$, $\beta=a-i b$ and replace in the expression. After simplifications using the conjugates, $$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac 1{\alpha}+\frac 1{\beta})+3{\alpha}{\beta}=\frac{4 (a+1) a}{a^2+b^2}+3 a^2+3 b^2-2$$ where $a=1$ and $b=\frac 1{\sqrt 3}$.

Answer 3

This is a quadratic polynomial with real coefficients and a negative discriminant [ $ \ (-6)^2 \ - \ 4 \ \cdot \ 3 \ \cdot \ 4 \ < \ 0 \ $ ] so its roots are the complex conjugates $ \ z \ $ and $ \ \overline{z} \ $ . Using Viete's formulas, much as André Nicolas has done, on $ \ 3 \ ( \ x^2 \ - \ 2 x \ + \ \frac{4}{3} \ ) \ $ , we have $ \ z \ \overline{z} \ = \frac{4}{3} \ $ . This gives us the third term as just $ \ 3 \ z \ \overline{z} \ = \ 4 \ $ . Further, the modulus of either root is found from $ \ z \ \overline{z} \ = \ | z |^2 \ = \ \frac{4}{3} \ \ \Rightarrow \ \ | z | \ = \ \frac{2}{\sqrt{3}} \ $ .

The sum of the roots is $ \ z \ + \ \overline{z} \ = \ 2 \ Re(z) \ = \ 2 \ \ \Rightarrow \ \ Re(z) \ = \ 1 \ $ . The factor in parentheses is $ \ \frac{1}{z} \ + \ \frac{1}{\overline{z}} \ = \ \frac{z \ + \ \overline{z}}{z \ \overline{z}} $ . Hence, the second term in the expression is $ \ 2 \ \left( \frac{2 Re(z)}{| z |^2 } \right) \ = \ \frac{4}{\frac{4}{3}} \ = \ 3 \ $ .

It remains to calculate the first term , $ \ \frac{z}{\overline{z}} \ + \ \frac{\overline{z}}{z} \ = \ \frac{z^2 \ + \ \overline{z}^2 }{z \ \overline{z}} \ $ . We have established that, say, $ \ z \ = \ 1 \ + \ ib \ $ , so the square of its modulus is $ \ 1 \ + \ b^2 \ = \ \frac{4}{3} \ \ \Rightarrow \ \ b^2 \ = \ \frac{1}{3} \ $ . But the numerator of this first term is $$ \ z^2 \ + \ \overline{z}^2 \ = \ 2 \ \cdot \ 1 \ - \ 2 \ b^2 \ \ = \ 2 \ - \ \frac{2}{3} \ = \ \frac{4}{3} \ \ , $$ so the value of the first term is $ \ \frac{\frac{4}{3}}{\frac{4}{3}} \ = \ 1 \ . $

We conclude that the sum of the three terms is $ \ 1 \ + \ 3 \ + \ 4 \ = \ 8 \ $ and that the roots are $ \ 1 \ \pm \ i\frac{1}{\sqrt{3}} \ $ .