Q)  If 𝛼, β are zeroes of quadratic polynomial x2 + x – 2, find the value of \frac{\alpha}{\beta} + \frac{\beta}{\alpha}

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

Therefore, x2 + x – 2 = 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}

∴    α + β =  \frac{(- 1)}{1} = – 1 …………(i)

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

∴  α . β = \frac{(- 2)}{1}  = – 2 ………… (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{(- 1)^2 - 2 (- 2)}{- 2}

= \frac{ 1 + 4}{- 2}

= - \frac{5}{2}

Therefore, the value of (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}) is – \frac{5}{2}

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