If 𝛼, β are zeroes of quadratic polynomial f(x) = 6x 2 + 11x – 10, find the value of 𝛼/𝛽 + 𝛽 /𝛼.

Question

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

Sapling Academy · Accepted Answer

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) =  $\frac{-b}{a}$

$	herefore$   α + β =  $\frac{(- 11)}{6}$ ............(i)

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

$	herefore$ 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:

$	herefore \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)

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

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

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

= $\frac{141 	imes 6}{36 	imes (- 10)}$ = - $\frac{141}{60}$

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

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