Question

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6. If a.ß are the zeroes of the polynomial 3x^2 + 8x+2, then 1/a +1/ß
is

Answer 1

Answer :

The required value is -4

Given :

The zeroes of the polynomial

3x² + 8x + 2 are :

$\alpha \: \: and \: \: \beta$

Concept to be used :

$\bullet \: \: \rm Sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\bullet \: \: \rm Product \: of \: the \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

To Find :

The value of

$\rm \dfrac{1}{\alpha}+\dfrac{1}{\beta}$

Solution :

The quadratic polynomial is

3x² + 8x + 2

From the sum relation we have ,

$\rm \implies \alpha + \beta = -\dfrac{8}{3} \longrightarrow (1)$

Now from product relation,

$\rm \implies \alpha \beta=\dfrac{2}{3} \longrightarrow(2)$

Dividing (1) by (2) we have,

$\rm \implies \dfrac{\alpha + \beta}{\alpha\beta}=\dfrac{-\dfrac{8}{3}}{\dfrac{2}{3}}\\\\ \implies \dfrac{\alpha}{\alpha\beta}+\dfrac{\beta}{\alpha\beta}=-\dfrac{8}{2}\\\\ \implies \dfrac{1}{\alpha}+\dfrac{1}{\beta}= -4$

Answer 2

Given:-

Polynomial 3x² + 8x + 2 whose zeros are $\alpha$ and $\beta$.

To find:-

${\sf{ \frac{1}{ \alpha } + \frac{1}{ \beta }}}$

Solution:-

In a polynomial ax² + bx + c = 0 :-

${\sf{\boxed{Sum \: of \: zeros = \frac{ - b}{a}}}} \\ {\sf{\boxed{Product \: of \: zeros = \frac{c}{a} }}}$

Here a = 3, b = 8 and c = 2

${\sf{\alpha + \beta = \frac{ - 8}{3} ..........(i)}} \\ {\sf{\alpha \beta = \frac{2}{3} .........(ii) }}$

Dividing (i) by (ii) we have,

$↦{\sf{\dfrac{\alpha + \beta}{\alpha\beta}=\dfrac{\dfrac{-8}{3}}{\dfrac{2}{3}}}}\\↦ {\sf{\dfrac{\alpha}{\alpha\beta}+\dfrac{\beta}{\alpha\beta}=-\dfrac{8}{2} }}\\↦{\sf{\frac{1}{\alpha} + \frac{1}{\beta}= - 4}}$

${\therefore} {\sf{\frac{1}{\alpha} + \frac{1}{\beta}= - 4}}$