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If alpha and beta are the zeroes of $3 x^{2} - 6 x + 4$ , find the value of $\frac{a lp ha}{b e t a} + \frac{b e t a}{a lp ha}$ .

Class 10Zeroes and CoefficientsAlgebra — Polynomials

A1
B0
C $\frac{3}{2}$
D2

Step-by-Step Solution

Sum =

\frac{6}{3} = 2

, product =

\frac{4}{3} . \frac{a lp ha}{b e t a} + \frac{b e t a}{a lp ha} = \frac{a lp h a ^{2} + b e t a ^{2}}{a lp ha \cdot b e t a} = \frac{( s u m ) ^{2} - 2 \cdot p ro d u c t}{p ro d u c t} = \frac{4 - \frac{8}{3}}{\frac{4}{3}} = \frac{4/3}{4/3} = 1

.

Answer: A) 1

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