Question

QUESTION:
If $\alpha$ & $\beta$ are the zeroes of the polynomial $f(x) = ax^2 + bx + c$, then find a polynomial whose zeroes are $\alpha + \dfrac{1}{\beta}$ & $\beta + \dfrac{1}{\alpha}$.
Help please :) ​

Answer 1

Answer:-

If α and β are the zeroes of the quadratic polynomial, then

aα

2

+bα+c=0 and

aβ

2

+bβ+c=0

or, aα

2

+bα − aβ

2

−bβ=0

Then, α+β=

a

−b

and αβ=

ab

c

Answer 2

Since $\alpha$ and $\beta$ are the zeroes of $ax^2 + bx + c$, thus:

$\alpha + \beta = - \dfrac{b}{a}$ &

$\alpha \beta = \dfrac{c}{a}$

Now, zeroes of the required polynomial are $\alpha + \dfrac{1}{\beta}$ & $\beta + \dfrac{1}{\alpha}$.

∴ Sum = $\alpha + \dfrac{1}{\beta} + \beta + \dfrac{1}{\alpha}$

= $\alpha + \beta + \dfrac{\alpha + \beta}{\alpha \beta}$

= $\dfrac{- b}{a} + \dfrac{\frac{-b}{a}}{\frac{c}{a}}$

= $- \dfrac{ b}{a} - \dfrac{b}{c}$

= $\dfrac{- bc - ab}{ac}$

= $\dfrac{- b(a + c)}{ac}$

Now, product = $(\alpha + \dfrac{1}{\beta}) (\beta + \dfrac{1}{\alpha})$

= $\alpha \beta + 1 + 1 + \dfrac{1}{\alpha \beta}$

= $\dfrac{c}{a} + \dfrac{a}{c} + 2$

= $\dfrac{c^2 + a^2 + 2ac}{ac}$

= $\dfrac{(a + c)^2}{ac}$

Therefore, required polynomial:

$k[x^2 - Ax + B]$ (A = sum of zeroes, B = product.

= $k[x^2 - (\dfrac{- b(a + c)}{ac})x + \dfrac{(a + c)^2}{ac}]$

= $k[x^2 + \dfrac{b(a + c)}{ac} + \dfrac{(a + c)^2}{ac}]$. _(Ans.)