Question

If alpha and bita are the zeros of polynomial 3x2-2x-7 then find the value of alpha/ bita+ bita/alpha​

Answer 1

Answer:

this is my answer. if you like give thanks. My writing is not that good, but it is good .

Answer 2

Answer:

༒ Given ➽

α and β are the zeros of the polynomial

$\sf p(s) = 3 {s}^{2} - 6s + 4$

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༒ To Calculate ➽

$\sf \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } + 2( \frac{1}{ \alpha } + \frac{1}{ \beta } ) +3\alpha\beta$

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༒ Key point ➽

For any quadratic polynomial

$\bf a {x}^{2} + bx + c \\ \\ \sf Sum \: of \: zeros = \frac{ - b}{a} \\ \\ \sf Product \: of \: zeros = \frac{c}{a}$

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༒ Solution ➽

Here,

α + β = 2

and

$\Large \bf \alpha \beta = \frac{4}{3}$

Now,

$\sf \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } + 2( \frac{1}{ \alpha } + \frac{1}{ \beta } ) +3\alpha\beta\\ \\ \sf = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } + 2( \frac{ \beta + \alpha }{ \alpha \beta } )+3\alpha\beta\\ \\ = \sf \frac{ { \alpha }^{2} + \beta {}^{2} + 2( \alpha + \beta ) }{ \alpha \beta } +3\alpha\beta\\ \\ \sf = \frac{ {( \alpha + \beta ) }^{2} - 2 \alpha \beta + 2( \alpha + \beta )}{ \alpha \beta }+3\alpha\beta \\ \{ \bf add \: and \: subtract \: 2 \alpha \beta \: in \: numerator \}$

$\dfrac{ {2}^{2} - 2( \dfrac{4}{3} ) + 2(2)}{ \dfrac{4}{3} } +3\times\frac{4}{3}\\ \\ \sf = \dfrac{4 - \dfrac{8}{3} + 4}{ \dfrac{4}{3} }+4$

$\sf = \dfrac{ \dfrac{12 - 8 + 12}{ \cancel{3}} }{ \dfrac{4}{ \cancel{3}} } +4\\ \\ \sf = \frac{16}{4} +4\\ \\ \Large \purple{ = \bf8}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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