Let $a, b, c, p, q$ be real numbers. Suppose $\alpha, \beta$ are the roots of the equation $x^{2}+2

Question

Let $a, b, c, p, q$ be real numbers. Suppose $\alpha, \beta$ are the roots of the equation $x^{2}+2 p x+q=0$ and $\alpha, 1 / \beta$ are the roots of the equation $a x^{2}+2 b x+c=0$, where $\beta^{2} 
otin\{-1,0,1\}$. Statement $-1:\left(p^{2}-qight)\left(b^{2}-a cight) \geq 0$ Statement - $2: b 
eq p a$ or $c 
eq q a$

Professor PC Thomas Classes · Answer

(b) : The two equations have roots represented as $\alpha, \beta$ and $\alpha, \frac{1}{\beta}$. One root is\ seen to be common. .\ If these equations have both roots common, then $\beta=\frac{1}{\beta}$ which means\ $\beta^{2}=1$ but $\beta^{2} 
otin\{-1,0,1\}$\ Suppose roots are imaginary, then $\beta=\bar{\alpha}$\ Also $\frac{1}{\beta}=\bar{\alpha}$. But then $\beta=\frac{1}{\beta} \Rightarrow \beta^{2}=1$\ (not possible) Thus roots of both equations are real which gives\ $\left(p^{2}-qight)\left(b^{2}-a cight) \geq 0$\ Again $\alpha+\frac{1}{\beta} 
eq \alpha+\beta \Rightarrow-\frac{2 b}{a} 
eq-2 p \Rightarrow b 
eq p a$\ Similarly, $\alpha 	imes \frac{1}{\beta} 
eq \alpha \beta \Rightarrow \frac{c}{a} 
eq q \Rightarrow c 
eq q a$\ Where we have used formula for product of roots and sum of roots.\ Thus both statement - 1 and statement - 2 are true but statement - 2 doesn't explain the statement - 1 .\