Question

if alpha and beta are the zeroes of polynomial f(x)=4x^2-5x+1 find a quadratic polynomial whose zeroes are alpha^2/beta and beta^2/alpha

Answer 1

Answer:

$\boxed{\huge{\pink{16 {x}^{2} - 65x + 4}}}$

Step-by-step explanation:

$\bold{\green{Given}}\longrightarrow \\ \alpha \: and \: \beta \: are \: zeroes \: of \: polynomial \\ f(x) = 4 {x}^{2} - 5x + 1$

$\bold{\red{To \: find}}\longrightarrow \\ a \: quadratic \: polynomial \: whose \: zeroes \: are \: \\ \dfrac{ { \alpha }^{2} }{ \beta } \: and \dfrac{ { \beta }^{2} }{ \alpha }$

$\bold{\blue{Concept \: used}}\longrightarrow \\ polynomial \: is \: \\ {x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes$

$\bold{\pink{Solution}}\longrightarrow \\ f(x) = 4 {x}^{2} - 5x + 1 \\ = 4 {x}^{2} - (4 + 1)x + 1 \\ = 4 {x}^{2} - 4x - x + 1 \\ = 4x(x - 1) - 1(x - 1) \\ = (x - 1) \: (4x - 1) \\ if \: x - 1 = 0 \: \: \: = > x = 1 \\ if \: 4x - 1 = 0 = > x = \dfrac{1}{4} \\ so \\ \alpha = 1 \: and \: \beta = \dfrac{1}{4} \\ \dfrac{ { \alpha }^{2} }{ \beta } = \dfrac{ {(1)}^{2} }{ \dfrac{1}{4} } \\ \dfrac{ { \alpha }^{2} }{ \beta } = 4 \\ \dfrac{ { \beta }^{2} }{ \alpha } = \dfrac{( \dfrac{ {1} }{4}) ^{2} }{1} \\ \dfrac{ { \beta }^{2} }{ \alpha } = \dfrac{1}{16} \\ now$

$sum \: of \: zeroes \: = \dfrac{ { \alpha }^{2} }{ \beta } + \dfrac{ { \beta }^{2} }{ \alpha } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4 + \dfrac{1}{16} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{65}{16}$

$product \: of \: zeroes = \dfrac{ { \alpha }^{2} }{ \beta } \: \: \: \dfrac{ { \beta }^{2} }{ \alpha } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \alpha \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{1}{4}$

$required \: polynomial \: is \\ k( {x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes \: ) \\ = k( {x}^{2} - ( \dfrac{65}{16} )x + \dfrac{1}{4} ) \\ = k( \dfrac{16 {x}^{2} - 65x + 4}{16} ) \\ taking \: k \: = 16 \: we \: get \\ = 16( \dfrac{16 {x}^{2} - 65x + 4}{16} ) \\ = \pink{16 {x}^{2} - 65x + 4}$