Question

If Alpha n beta are two zeros of polynomial 3x²-4x+1, fibd quadratic polynomial whose zeros are alpha square/beta + beta square/alpha​

Answer 1

Step-by-step explanation:

Gven :

$\alpha \: \: and \: \: \beta$

are the zeros of the polynomial

$3 {x}^{2} - 4x + 1$

To find :

The quadratic polynomial whose zeros are ;

$\frac{ { \alpha }^{2} }{ \beta } \: \: and \: \: \frac{ { \beta }^{2} }{ \alpha }$

Solution :

For the given quadratic polynomial

$3 {x}^{2} - 4x + 1$

Sum of the roots =

$\alpha \: + \: \beta = - \frac{ - 4}{3} = \frac{4}{3}$

Product of the roots =

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

Now,

$\frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha } = \frac{ { \alpha }^{3} + { \beta }^{3} }{ \alpha \beta } \\ = \frac{( \alpha + \beta )( {( \alpha + \beta )}^{2} - 3\alpha \beta )}{ \alpha \beta }$

=

$\frac{ \frac{4}{3} ( { \frac{4}{3} }^{2} - 3 \times \frac{1}{3} )}{ \frac{1}{3} } = 4( \frac{16}{9} - 1) = \frac{28}{9}$

and,

$\frac{ { \alpha }^{2} }{ \beta } \times \frac{ { \beta }^{2} }{ \alpha } = \alpha \beta = \frac{1}{3}$

Required quadratic polynomial =

${x}^{2} - sx + p$

where s = sum of zeros, and p = product of zeros

therefore,

${x}^{2} - \frac{28}{9}x \: + \frac{1}{3} \\ = \frac{1}{9} (9 {x}^{2} - 28x + 3)$

Hence, the required polynomial.