Theory Of Equations Ques 6
- The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation, $6 x^{2}-11 x+\alpha=0$ are rational numbers is
(2019 Main, 9 Jan II)
(a) 5
(b) 2
(c) 4
(d) 3
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Answer:
Correct Answer: 6.(d)
Solution:
- For the roots of quadratic equation $a x^{2}+b x+c=0$ to be rational $D=\left(b^{2}-4 a c\right)$ should be perfect square.
In the equation $6 x^{2}-11 x+\alpha=0$
$ a=6, b=-11 \text { and } c=\alpha $
$\therefore$ For roots to be rational
$D=(-11)^{2}-4(6)(\alpha)$ should be a perfect square.
$\Rightarrow D(\alpha)=121-24 \alpha$ should be a perfect square Now,
$D(1)=121-24=97$ is not a perfect square.
$D(2)=121-24 \times 2=73$ is not a perfect square.
$D(3)=121-24 \times 3=49$ is a perfect square.
$D(4)=121-24 \times 4=25$ is a perfect square.
$D(5)=121-24 \times 5=1$ is a perfect square.
and for $\alpha \geq 6, D(\alpha)<0$, hence imaginary roots.
$\therefore$ For 3 values of $\alpha(\alpha=3,4,5)$, the roots are rational.
BETA
