Q)  If 𝛼, β are zeroes of quadratic polynomial f(x) = 6 x2 + 11 x – 10, find the value of \frac{\alpha}{\beta} + \frac{\beta}{\alpha}

Ans: In the given polynomial equation,  to find zeroes, we will start with f(x) = 0

Therefore, 6 x2 + 11 x – 10 = 0

Step 1: Given that the roots of the polynomial are α and β.

We know that sum of roots (α + β) =  \frac{-b}{a}

\therefore   α + β =  \frac{(- 11)}{6} …………(i)

Next, we know that the product of the roots (α x β) = \frac{c}{a}

\therefore α . β = \frac{(- 10)}{6} ………… (ii)

Step 2: Next, we need to find the value of \frac{\alpha}{\beta} + \frac{\beta}{\alpha}

Let’s solve this to simplify:

\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha ^2 + \beta ^2}{\alpha \beta} ………(iii)

We know that (a + b)2 = a2 + b2 + 2 a b

or we can say that a2 + b2 = (a + b)2 – 2 a b

Therefore, α2 + β2 = (α + β)2 – 2 α β

Transferring this value in equation (iii), we get:

\therefore \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha + \beta)^2 - 2 \alpha \beta }{\alpha \beta}

Step 3: Next, we transfer values of (α + β) and α β from equations (i) and (ii)

\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha + \beta)^2 - 2 \alpha \beta }{\alpha \beta}

= \frac{(\frac{-11}{6})^2 - 2 (\frac{ - 10}{6})}{\frac{- 10}{6}} = \frac{\frac{121}{36} + \frac{20}{6}}{\frac{- 10}{6}}

= \frac{\frac{121 + 120}{36}}{\frac{- 10}{6}} = \frac{\frac{141}{36}}{\frac{- 10}{6}}

= \frac{141 \times 6}{36 \times (- 10)} = – \frac{141}{60}

Therefore, the value of (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}) is – \frac{141}{60}

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