Question

If alpha and bita are the zeros of the quadratic
polynomial p(s) = 3s² – 6s +4, find the value
Alpha/bita+bita/alpha+2(1/alpha+1/bita)+3alpha*bita​

Answer 1

༒ Given ➽

α and β are the zeros of the polynomial

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

___________________________

༒ To Calculate ➽

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

___________________________

༒ Key point ➽

For any quadratic polynomial

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

___________________________

༒ Solution ➽

Here,

α + β = 2

and

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

Now,

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

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

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

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

___________________________

Answer 2

Answer:

Given :-

• If α and β are the zeros of the quadratic polynomial p(s) = 3s² - 6s + 4.

To Find :-

• What is the value of α/β + β/α + 2(1/α + 1/β) + 3αβ.

Solution :-

Given Equation :

$\bigstar\: \: \sf\bold{3s^2 - 6s + 4}$

where,

• a = 3
• b = - 6
• c = 4

First, we have to find the sum and product of roots :

${\small{\bold{\purple{\underline{\leadsto\: In\: case\: of\: sum\: of\: roots\: :-}}}}}$

As we know that,

$\clubsuit$ Sum of roots Formula :

$\mapsto\: \sf\boxed{\bold{\pink{Sum\: of\: roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

We have :

• a = 3
• b = - 6

According to the question by using the formula we get,

$\implies \sf \alpha + \beta =\: \dfrac{- (- 6)}{3}$

$\implies \sf \alpha + \beta =\: \dfrac{\cancel{6}}{\cancel{3}}$

$\implies \sf\bold{\green{\alpha + \beta =\: 2}}$

${\small{\bold{\purple{\underline{\leadsto\: In\: case\: of\: product\: of\: roots\: :-}}}}}$

As we know that,

$\clubsuit$ Product of roots Formula :

$\mapsto \sf\boxed{\bold{\pink{Product\: of\: roots\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

We have :

• a = 3
• c = 4

According to the question by using the formula we get,

$\implies \sf \bold{\green{\alpha\beta =\: \dfrac{4}{3}}}$

Now, we have to find the value of :

$\dashrightarrow \sf\bold{\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} + 2\bigg(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\bigg) + 3\alpha\beta}$

We have :

$\bullet\: \: \rm{\bold{\alpha + \beta =\: 2}}$

$\bullet\: \: \rm{\bold{\alpha\beta =\: \dfrac{4}{3}}}$

According to the question,

$\longrightarrow \sf \dfrac{\alpha^2 + \beta^2}{\alpha\beta} + 2\bigg(\dfrac{\beta + \alpha}{\alpha\beta}\bigg) + 3\alpha\beta$

$\longrightarrow \sf \dfrac{\alpha^2 + \beta^2 + 2(\alpha + \beta)}{\alpha\beta} + 3\alpha\beta$

$\longrightarrow \sf \dfrac{(\alpha + \beta)^2 - 2\alpha\beta + 2(\alpha + \beta)}{\alpha\beta} + 3\alpha\beta\: \: \bigg\lgroup \sf\bold{\pink{a^2 + b^2 =\: (a + b)^2 - 2ab}}\bigg\rgroup$

$\longrightarrow \sf \dfrac{(2)^2 - 2\bigg(\dfrac{4}{3}\bigg) + 2(2)}{\dfrac{4}{3}} + 3\bigg(\dfrac{4}{3}\bigg)$

$\longrightarrow \sf \dfrac{4 - \dfrac{8}{3} + 4}{\dfrac{4}{3}} + \dfrac{\cancel{12}}{\cancel{3}}$

$\longrightarrow \sf \dfrac{\dfrac{12 - 8 + 12}{3}}{\dfrac{4}{3}} + 4$

$\longrightarrow \sf \dfrac{\dfrac{16}{3}}{\dfrac{4}{3}} + 4$

$\longrightarrow \sf \dfrac{16}{3} \times \dfrac{3}{4} + 4$

$\longrightarrow \sf \dfrac{\cancel{48}}{\cancel{12}} + 4$

$\longrightarrow \sf 4 + 4$

$\longrightarrow \sf\bold{\red{8}}$

$\therefore \small{\bf{\underline{\underline{The\: value\: of\: \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} + 2\bigg(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\bigg) + 3\alpha\beta\: is\: 8\: .}}}}$