Question

if alpha and beeta are zeros of polynomial 6x^2-7x-3 then form a quadratic polynomial whose zeros are 1/a and 1/b​

Answer 1

❖❖❖❖❖ Question ❖❖❖❖❖

$if \: \alpha \: and \: \beta \: are \: \\ zeroes \: of \: polynomial \: 6 {x}^{2} - 7x - 3 \\ \: then \: form \: a \: quadratic \: polynomial \: \\ whose \: zeroes \: are \: \frac{1}{ \alpha } and \: \frac{1}{ \beta } .$

❖❖❖❖❖Solution❖❖❖❖❖

Let us firstly find the zeroes of polynomial

$6 {x}^{2} - 7x - 3$

by splitting the middle term

$6 {x}^{2} - 7x - 3 = 0 \\ \\ 6 {x}^{2} - 9x + 2x - 3 = 0 \\ \\ 3x(2x - 3) + 1(2x - 3) = 0 \\ \\ (3x + 1)( 2x - 3) = 0 \\ \\ so \\ \\ x \: = \frac{ - 1}{3} \\ and \\ x = \frac{3}{2} \\ are \: two \: zeroes \: of \: given \: polynomial \\ \\ \alpha = \frac{ - 1}{3} \\ and \\ \beta = \frac{3}{2}$

Now,

we have to form a quadratic polynomial whose zeroes are

$\frac{1}{ \alpha } = \frac{3}{ - 1} = - 3 \\ \\ and \\ \\ \frac{1}{ \beta } = \frac{2}{3}$

Let,

we have to form a quadratic polynomial

$a {x}^{2} + bx + c$

whose zeroes are -3 and 2/3

so,

as we know

$sum \: of \: zeroes = \frac{ - b}{a} \\ \\ - 3 + \frac{2}{3} = \frac{ - b}{a} \\ \\ \frac{ - 9 + 2}{3} = \frac{ - b}{a} \\ \\ \frac{ - 7}{3} = \frac{ - b}{a} \: \: \: \: - - - eqn(1)$

and

$product \: of \: zeroes \: = \frac{c}{a} \\ \\ - 3 \times \frac{2}{3} = \frac{c}{a} \\ \\ \frac{ - 6}{3} = \frac{c}{a} \: \: \: \: - - - eqn(2)$

By comparing eqn(1) and (2) we will get,

a = 3 , b = 7 , c = - 6

Hence the polynomial formed will be

$3 {x}^{2} + 7x - 6$