Question

If α and β are the zeroes of the polynomial 6y 2 − 7y + 2, find a quadratic polynomial whose zeroes are 1 α and 1 β .

Answer 1

Answer:

$2y^2-7y+6=0$

Step-by-step explanation:

We are given that $\alpha$ and $\beta$ are the zeroes of the polynomial $6y^2-7y+2$

$y^2-\frac{7}{6}y+\frac{1}{3}$

We have to find a quadratic polynomial whose zeroes are $1/\alpha$ and $1/\beta$.

General quadratic equation

$x^2-(sum\;of\;zeroes)x+ product\;of\;zeroes$

We get

$\alpha+\beta=\frac{7}{6}$

$\alpha \beta=\frac{1}{3}$

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

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7/6}{1/3}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7}{6}\times 3=7/2$

$\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{\alpha \beta}$

$\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{1/3}=3$

Substitute the values

$y^2-(7/2)y+3=0$

$2y^2-7y+6=0$

Hence, the quadratic polynomial whose zeroes are $1/\alpha$ and $1/\beta$ is given by

$2y^2-7y+6=0$