Question

Given that one of the zeroes of the cubic polynomial
ax3 + bx2 + cx + d is zero, the sum of the reciprocal of
the other two zeroes is
a) b/a
b) b/c
c) -b/a
d) -b/c​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{One\;of\;zeroes\;of\;ax^3+bx^2+cx+d\;is\;0}$

$\underline{\textbf{To find:}}$

$\textsf{The sum of the reciprocal of the other two zeroes}$

$\underline{\textbf{Solution:}}$

$\mathsf{Let\;the\;other\;two\;zeroes\;be\;\alpha\;and\;\beta}$

$\mathsf{Then,}$

$\mathsf{\sum_1=\alpha+\beta+0=\dfrac{-b}{a}}$

$\implies\mathsf{\alpha+\beta=\dfrac{-b}{a}}$

$\mathsf{\sum_2=\alpha\beta+\beta\,0+0\,\alpha=\dfrac{c}{a}}$

$\implies\mathsf{\alpha\beta=\dfrac{c}{a}}$

$\mathsf{Now,}$

$\textsf{Sum of the reciprocal of other two zeroes}$

$\mathsf{=\dfrac{1}{\alpha}+\dfrac{1}{\beta}}$

$\mathsf{=\dfrac{\alpha+\beta}{\alpha\beta}}$

$\mathsf{=\dfrac{\dfrac{-b}{a}}{\dfrac{c}{a}}}$

$\mathsf{=\dfrac{-b}{c}}$

$\underline{\textbf{Answer:}}$

$\textsf{Option(d) is correct}$

$\underline{\textbf{Find more:}}$

If the zeroes of cubic polynomial x3-15x2+74x-120 are the form a,a+b and a+2b for some positive real number, than the value of a and b are​

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Answer 2

Answer:

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