Question

If alpha and Beta are the zeroes of polynomial p(x) = ax² + bx+c (a is not equal to 0, c is not equal to 0), find the quadratic polynomial with zeroes (-1/alpha) and (-1/beta).​

Answer 1

Step-by-step explanation:

Given:

$p(x) = a {x}^{2} + bx + c$

where a and c≠0.

To find: Find the quadratic polynomial with zeroes $\frac{-1}{\alpha}\: and\:\frac{-1}{\beta}$.

Solution:

If $\alpha\:and\:\beta$are zeros of given polynomial ,then

$\alpha + \beta = \frac{ - b}{a} ...eq1 \\ \\ \alpha \beta = \frac{c}{a} ...eq2$

To find the polynomial having zeros $\frac{-1}{\alpha}\: and\:\frac{-1}{\beta}$.

Divide eq1 by eq2

$\frac{ \alpha + \beta }{ \alpha \beta } = \frac{ \frac{ - b}{a} }{ \frac{c}{a} } \\ \\ or \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ - b}{c}...eq3$

multiply eq3 by (-)

$- \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{b}{c}$

and

$\left( \frac{ - 1}{ \alpha } \right)\left( \frac{ - 1}{ \beta} \right) = \frac{a}{c} \\ \\ \frac{1}{ \alpha \beta } = \frac{a}{c}$

The polynomial is

${x}^{2} - \left( - \frac{1}{ \alpha } - \frac{1}{ \beta } \right )x + \frac{1}{ \alpha \beta }$

or

${x}^{2} - \frac{b}{c} x + \frac{a}{c}$

Take LCM and equate the polynomial to zero for simplification

$\frac{c {x}^{2} - bx + a }{c} = 0 \\ \\ or \\ \\ c {x}^{2} - bx + a = 0 \:$

Thus, the quadratic polynomial is

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

Final answer:

The quadratic polynomial with zeroes $\frac{-1}{\alpha}\: and\:\frac{-1}{\beta}$ is $\bold{\pink{c {x}^{2} - bx + a}}$.

Hope it helps you.

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