Let α and β be the roots of X²+7X+12=0.Then the value of ≤ft(\;{α²}{β}+\;{β²}{α}) will be

Correct Answer is : − 9112-\;91/12−1291

Correct Answer is : − 9112-\;91/12−1291

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Question : 34 of 39

Marks: +1, -0

Let

\alpha

\beta

X^{2}+7X+12=0

\left(\;\frac{\alpha^{2}}{\beta}+\;\frac{\beta^{2}}{\alpha}\right)

will be

Solution:

Concept:

Use sum and product of roots to evaluate the expression without solving the quadratic directly.

Explanation:

Given

X^2+7X+12=0

, roots

\alpha

\beta

Sum:

\alpha+\beta = -7

, product:

\alpha\beta = 12

Expression:

\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} = \frac{\alpha^3+\beta^3}{\alpha\beta}

Now

\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)

Substitute:

(-7)^3 - 3\cdot12\cdot(-7) = -343 + 252 = -91

Divide by

\alpha\beta=12

-\frac{91}{12}

Answer:

-\frac{91}{12}

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