If alpha and beta are the zeros of the polynomial y 2 - 7y +2. find th

1.	If alpha and beta are the zeros of the polynomial y 2 - 7y +2. find the quadratic polynomial whose zeros are a1/alpha and 1/beta
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{2y^{2}-7y+1=0}}}\\$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{ \underline \bold{<klux>GIVEN</klux>:}} \\ \tt: \implies {y}^{2}-7y+2= 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: zeroes \\ \\ \red{ \underline \bold{To \: Find:}} \\ \tt: \implies Polynomial\:whose\:zeroes\:are\:\frac{1}{\alpha}\:and\:\frac{1}{\beta} = ?$ • ACCORDING to given QUESTION : $\tt \circ \: {y}^{2} -7y +2 = 0 \\ \\ \tt \circ \: \alpha + \beta = \frac{ - b}{a} = {7} \\ \\ \tt \circ \: \alpha \beta = \frac{c}{a} = 2 \\ \\ \bold{For \: sum \: of \: zeroes}$ $\tt: \implies \frac{1}{ \alpha } + \frac{1}{ \beta } \\ \\ \tt: \implies \frac{ \beta + \alpha }{ \alpha \beta } \\ \\ \green{\tt: \implies \frac{7}{2} } \\ \\ \green{\tt \therefore \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{7}{2} } \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies \frac{1}{ \alpha } \times \frac{1}{ \beta } \\ \\ \tt: \implies \frac{1}{ \alpha \beta } \\ \\ \tt: \green{\implies \frac{1}{2} } \\ \\ \green{\tt \therefore \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{2}} \\ \\ \bold{For \: Quadratic \: eqn} \\ \tt: \implies {y}^{2} - {(sum \: of \: zeroes})y + ( product \: of \: zeroes) = 0 \\ \\ \tt: \implies {y}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )y+ \frac{1}({ \alpha } \times \frac{1}{ \beta }) = 0 \\ \\ \tt: \implies {y}^{2} - \frac{7}{2}y + \frac{1}{2} = 0 \\ \\ \tt: \implies \frac{2 {y}^{2} - 7y + 1 }{2} = 0 \\ \\ \green{\tt: \implies 2 {y}^{2} - 7y + 1 = 0}$

If alpha and beta are the zeros of the polynomial y 2 - 7y +2. find the quadratic polynomial whose zeros are a1/alpha and 1/beta

Answer»

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

$\green{\tt{\therefore{2y^{2}-7y+1=0}}}\\$

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

$\green{ \underline \bold{<klux>GIVEN</klux>:}} \\ \tt: \implies {y}^{2}-7y+2= 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: zeroes \\ \\ \red{ \underline \bold{To \: Find:}} \\ \tt: \implies Polynomial\:whose\:zeroes\:are\:\frac{1}{\alpha}\:and\:\frac{1}{\beta} = ?$

• ACCORDING to given QUESTION :

$\tt \circ \: {y}^{2} -7y +2 = 0 \\ \\ \tt \circ \: \alpha + \beta = \frac{ - b}{a} = {7} \\ \\ \tt \circ \: \alpha \beta = \frac{c}{a} = 2 \\ \\ \bold{For \: sum \: of \: zeroes}$

$\tt: \implies \frac{1}{ \alpha } + \frac{1}{ \beta } \\ \\ \tt: \implies \frac{ \beta + \alpha }{ \alpha \beta } \\ \\ \green{\tt: \implies \frac{7}{2} } \\ \\ \green{\tt \therefore \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{7}{2} } \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies \frac{1}{ \alpha } \times \frac{1}{ \beta } \\ \\ \tt: \implies \frac{1}{ \alpha \beta } \\ \\ \tt: \green{\implies \frac{1}{2} } \\ \\ \green{\tt \therefore \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{2}} \\ \\ \bold{For \: Quadratic \: eqn} \\ \tt: \implies {y}^{2} - {(sum \: of \: zeroes})y + ( product \: of \: zeroes) = 0 \\ \\ \tt: \implies {y}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )y+ \frac{1}({ \alpha } \times \frac{1}{ \beta }) = 0 \\ \\ \tt: \implies {y}^{2} - \frac{7}{2}y + \frac{1}{2} = 0 \\ \\ \tt: \implies \frac{2 {y}^{2} - 7y + 1 }{2} = 0 \\ \\ \green{\tt: \implies 2 {y}^{2} - 7y + 1 = 0}$

If alpha and beta are the zeros of the polynomial y 2 - 7y +2. find the quadratic polynomial whose zeros are a1/alpha and 1/beta

• ACCORDING to given QUESTION :

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