Theory of Equations 5 Question 9
9. Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^{2}-x+\alpha=0$ has two distinct real roots $x _1$ and $x _2$ satisfying the inequality $\left|x _1-x _2\right|<1$. Which of the following interval(s) is/are a subset of $S$ ?
(a) $-\frac{1}{2},-\frac{1}{\sqrt{5}}$
(b) $-\frac{1}{\sqrt{5}}, 0$
(c) $0, \frac{1}{\sqrt{5}}$
(d) $\frac{1}{\sqrt{5}}, \frac{1}{2}$
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Answer:
Correct Answer: 9. (a,d)
Solution:
- Given, $x _1$ and $x _2$ are roots of $\alpha x^{2}-x+\alpha=0$.
$ \therefore \quad x _1+x _2=\frac{1}{\alpha} \text { and } x _1 x _2=1 $
Also, $\quad\left|x _1-x _2\right|<1$
$\Rightarrow \quad\left|x _1-x _2\right|^{2}<1 \quad \Rightarrow \quad\left(x _1-x _2\right)^{2}<1$
or $\quad\left(x _1+x _2\right)^{2}-4 x _1 x _2<1$
$\Rightarrow \quad \frac{1}{\alpha^{2}}-4<1$ or $\frac{1}{\alpha^{2}}<5$
$\Rightarrow \quad 5 \alpha^{2}-1>0$ or $(\sqrt{5} \alpha-1)(\sqrt{5} \alpha+1)>0$
$ \therefore \quad \alpha \in-\infty,-\frac{1}{\sqrt{5}} \cup \frac{1}{\sqrt{5}}, \infty $
Also, $\quad D>0$
$ \Rightarrow \quad 1-4 \alpha^{2}>0 \quad \text { or } \quad \alpha \in-\frac{1}{2}, \frac{1}{2} $
From Eqs. (i) and (ii), we get
$ \alpha \in-\frac{1}{2}, \frac{-1}{\sqrt{5}} \cup \frac{1}{\sqrt{5}}, \frac{1}{2} $
