Question

if alpha and bita are the zeroes of quadratic polynomial f(x)=6x2+x-2, then find 1/alpha + 1/bita ​

Answer 1

Step-by-step explanation:

f(x) = 6x² + x - 2

alpha & beta are the zeroes of above quadratic polynomial .

So , alpha + beta = -1/6

alpha×beta = -2/6 = -1/3

Now , 1/alpha + 1/beta

(alpha+beta)/(alpha*beta)

(-1/6)/(-1/3) = 1/6 × 3/1 = 3/6 = 1/2

Answer 2

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

α + β/αβ = 1/2

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$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• Polynomial : 6x² + x - 2 = 0

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To Find :

We have to find the value of $\sf{\frac{1}{\alpha} + \frac{1}{\beta}}$

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Solution :

As, it is given that

$\sf{\implies \frac{1}{\alpha} + \frac{1}{\beta}} \\ \\ \sf{\implies \frac{\alpha + \beta}{\alpha \beta}......(1)}$

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We know the formula to find the sum and product of zeroes.

$\large{\boxed{\sf{Sum \: of \: zeroes(\alpha + \beta) = \frac{-b}{a}}}}$

$\sf{\implies \alpha + \beta = \frac{-1}{6}......(2)}$

Now, product of zeroes

$\large{\boxed{\sf{Product \: of \: zeroes (\alpha \beta) = \frac{c}{a}}}}$

$\sf{\implies \alpha \beta = \frac{-2}{6}} \\ \\ \sf{\implies \alpha \beta = \frac{-1}{3}......(3)}$

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Now, From equation (1), (2) and (3)

We get,

$\sf{\implies \frac{\alpha + \beta}{\alpha \beta} = \dfrac{ \dfrac{ \dfrac{-1}{6} }{-1} }{3}} \\ \\ \sf{\implies \frac{\alpha + \beta }{\alpha \beta} = \frac{3}{6}} \\ \\ \sf{\implies \frac{\alpha + \beta}{\alpha \beta} = \frac{1}{2}}$