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4-2.Quadratic Equations and Inequations
hard
જો ${x^2} + px + 1$ એ સમીકરણ $a{x^3} + bx + c$ નો એક અવયવ હોય તો
A
${a^2} + {c^2} = - ab$
B
${a^2} - {c^2} = - ab$
C
${a^2} - {c^2} = ab$
D
એકપણ નહી.
(IIT-1980)
Solution
(c) Given that ${x^2} + px + 1$is factor of $a{x^3} + bx + c = 0$,
then let $a{x^3} + bx + c \equiv ({x^2} + px + 1)(ax + \lambda )$, where $\lambda $ is a constant. Then equating the coefficient of like powers of $x$ on both sides, we get
$0 = ap + \lambda ,\;\;b = p\lambda + a,\;c = \lambda $
$ \Rightarrow p = – \frac{\lambda }{a} = – \frac{c}{a}$
Hence $b = \left( { – \frac{c}{a}} \right)\,c + a$ or $ab = {a^2} – {c^2}$.
Standard 11
Mathematics
