Question

and beta are the zeros of the polynomial 3 x square - 4 x + 1 find a quadratic polynomial whose zeros are Alpha square by beta and beta square by Alpha​

Answer 1

$\huge \boxed{ \bf{question}}$


If alpha and beta are the zeros of the polynomial

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

Find a quadratic polynomial whose zeros are

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



$\huge \boxed{ \bf{answer}}$


Let

$\alpha \: and \: \beta$
be the zeroes of P(x)


A/Q,

$P(x) = 3 {x}^{2} - 4x + 1$

$= > \: 3 {x}^{2} - 4x + 1 = 0$

$= > 3 {x}^{2} - 3x - x + 1 = 0$

$= > 3x(x - 1) - 1(x - 1) = 0$

$= > (3x - 1)(x - 1) = 0$

$= > x = \frac{1}{3} \: or \: x = 1$


So,

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


Now, putting the values, we get

$\frac{ { \alpha }^{2} }{ \beta } = \frac{ { (\frac{1}{3} )}^{2} }{1} = \frac{ { (\frac{1}{9} )} }{1} = \frac{1}{9}$

$\frac{ { \beta }^{2} }{ \alpha } = \: \frac{ ({1})^{2}}{ \frac{1}{3} } = \frac{1}{ \frac{1}{3} } = 3$


So, the Zeroes of the other Quadratic Equation is

$\frac{1}{9} \: and \: 3$

Therefore, the required Quadratic Equation is :

$\boxed{ \bf{ {x}^{2} + \frac{1}{9} x + 3 }}$