Question

If the roots of the equation $- 3 < a < 3$ are real, then.

• A. $a < - 2$
• B. $x^{2} - 2ax + a^{2} + a - 3 = 0$
• C. $a < 2$
• D. $2 \leq a \leq 3$

Answer 1

Answer: (C)

$a < 2$

Answer 2

Roots of $ax^{2} + bx + c = 0$ are real. So$\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} =$

⇒ $\frac{2}{a}$⇒ $\frac{2}{b}$

⇒ $\frac{2}{c}$⇒ $- \frac{2}{a}$

Now we have two cases:

Case I : $ax^{2} + bx + c = 0$and $a(a + b) = 2bc$

⇒ $c(a + c) = 2ab$and $b(a + b) = 2ac$

Case II : $b(a + b) = ac$and $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$

⇒ $\alpha + \beta$and $\alpha,\beta$but it is impossible

Therefore, we get $x^{2} - 2x + 3 = 0$

Aliter : Students should note that the expression $\frac{1}{\alpha^{2}}$ will be less than or equal to zero if $\frac{1}{\beta^{2}}$ or otherwise $x^{2} + 2x + 1 = 0$.

Therefore$9x^{2} + 2x + 1 = 0$

i.e., $9x^{2} - 2x + 1 = 0$.