Question

If α and β are the zeros of 5x^2-x-2, Find a polynomial whose zeros are 3α^2 β and 3β^2α

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{125x^{2}+30x-72=0}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given:}} \\ \tt: \implies 5{x}^{2} -x - 2 = 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: zeroes \\ \\ \red{ \underline \bold{To \: Find:}} \\ \tt: \implies Polynomial\:whose\:zeroes\:are\:3\alpha^{2}\beta\:and\:3\beta^{2}\alpha = ?$

• According to given question :

$\tt \circ \: {5x}^{2} - x - 2 = 0 \\ \\ \tt \circ \: \alpha + \beta = \frac{ - b}{a} = \frac{1}{5} \\ \\ \tt \circ \: \alpha \beta = \frac{c}{a} = \frac{ - 2}{5} \\ \\ \bold{For \: sum \: of \: zeroes} \\ \tt: \implies {3 \alpha }^{2} \beta + 3 { \beta }^{2} \alpha \\ \\ \tt: \implies 3 \alpha \beta ( \alpha + \beta ) \\ \\ \tt: \implies 3 \times \frac{ - 2}{5} \times \frac{1}{5} \\ \\ \tt: \implies \frac{ - 6}{25} \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies {3 \alpha }^{2} \beta + 3 { \beta }^{2} \alpha \\ \\ \tt: \implies 9 {( \alpha \beta )}^{3} \\ \\ \tt: \implies 9 \times (\frac{ - 2}{5} )^{3} \\ \\ \tt: \implies 9 \times \frac{ - 8}{125} \\ \\ \green{\tt: \implies \frac{ - 72}{125} } \\ \\ \bold{For \: Quadratic\:eqn: } \\ \tt: \implies {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) = 0 \\ \\ \tt: \implies {x}^{2} - ( \frac{ - 6}{25} )x + ( \frac{ - 72}{125} ) = 0 \\ \\ \green{\tt: \implies 125 {x}^{2} + 30x - 72 = 0}$

Answer 2

$</p><p>\tt{\huge{\pink{Hello!!! }}}$

$\tt{ \purple{f(x_{1}) = 5 {x}^{2} - x - 2}}$

$\tt{ \blue{ \implies{ \alpha + \beta = \dfrac{ - b}{a} = \dfrac{1}{5} }}}$

$\tt{ \green{ \implies{ \alpha \beta = \dfrac{c}{a} = \dfrac{ - 2}{5} }}}$

New Zeroes are :

$3 { \alpha }^{2} \beta \: and \: 3 { \beta }^{2} \alpha$

$\tt{ \red{ \implies{sum \: of \: zeroes = 3 \alpha \beta ( \alpha + \beta ) = \frac{ - 6}{25} }}}$

$\tt{ \pink{ \implies{product \: of \: zeroes \: = \dfrac{-72}{625} }}}$

$</p><p>\tt{\green{f(x_{2}) = {x}^2 \:- \:(Sum\:Of\: Zeroes)\: +\: (Product\:Of\: Zeroes) }}}$

Hence the required Quadratic Equation becomes :

$</p><p>\tt{\blue{\implies{125 {x}^2 \: + \: 30 x \: - \: 72 \: = \: 0 }}}$