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4-2.Quadratic Equations and Inequations
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Let $a, b, c$ be non-zero real roots of the equation $x^3+a x^2+b x+c=0$. Then,
A
There are infinitely many such triples $a, b, c$
B
There is exactly one such triple $a, b, c$
C
There are exactly two such triples a, $b, c$
D
There are exactly three such triples a, $b, c$
(KVPY-2020)
Solution
(c)
We have,
$a b c =-c$
$a b+b c+c a =b$
$And\,\,a+b+c =-a \text { From Eq. (i), }$
$a b =-1(\text { since } c \neq 0 \text { ) }$
So, from Eq. $(iii)$, $c=-2 a+\frac{1}{a}$
and from Eq. $(ii)$ we get,
$\quad-1+2-\frac{1}{a^2}-2 a^2+1=-\frac{1}{2}$
$\Rightarrow \quad \quad 2 a^4-2 a^2-a+1=0$
$\Rightarrow \quad(a-1)\left(2 a^3+2 a^2-1\right)=0$
$\text { From here we get only two real and }$
$\text { non-zero values of } a, \text { hence there exists }$
$\text { two triplets of }(a, b, c)$
Standard 11
Mathematics
