Question

Let $a,b,c,p,q$ be real numbers. Suppose $\alpha ,\beta$ are the roots of the equation $x^2+2px+q=0$ and $\alpha ,\frac{1}{\beta }$ are the roots of the equation $a{x}^2+2bx+c=0$, where ${\beta }^2\notin \{-1,0,1\}$ .

STATEMENT 1 : $(p^2-q)(b^2-ac)\ge 0$
STATEMENT 2: $b\notin pa$ or $c\notin qa$.

• A. STATEMENT 1 is True, STATEMENT2 is True; STATEMENT 2 is a correct explanation for STATEMENT 1.
• B. STATEMENT 1 is True, STATEMENT 2 is True; STATEMENT 2 is NOT a correct explanation for STATEMENT 1.
• C. STATEMENT 1 is True, STATEMENT 2 is False.
• D. STATEMENT 1 is False, STATEMENT 2 is True.

Answer 1

The correct option is B STATEMENT 1 is True, STATEMENT 2 is True; STATEMENT 2 is NOT a correct explanation for STATEMENT 1.

Given $\alpha ,\beta$ are the roots of the equation
$x^2+2px+q=0$
$\Rightarrow \alpha +\beta =-2p$ ....(1)
and $\alpha \beta =q$ .....(2)
$D=4p^2-4q$ ...(3)
Also given $\alpha ,\frac{1}{\beta }$ are the roots of the equation $a{x}^2+2bx+c=0$
$\Rightarrow \alpha +\frac{1}{\beta }=-\frac{2b}{a}$ ....(4)
and $\frac{\alpha }{\beta }=\frac{c}{a}$ ....(5)
$D=4b^2-4ac$
Statement 1: $(p^2-q)(b^2-ac)\ge 0$
That mean we have to prove roots are real .
Let us suppose on the contrary that roots are imaginary .
Since, imaginary roots occurs in pair and they are complex conjugate of each other .
$\beta =\alpha$
And for the other equation $\beta =\frac{1}{\beta }$
$\Rightarrow {\beta }^2=1$
which is contradiction to ${\beta }^2\notin \{-1,0,1\}$
Hence, statement 1 is correct.
Now, let us assume on the contrary, $\beta =1$
So, from eqn (2) and eq (4), we get
$\alpha =q$ and $\alpha =\frac{c}{a}$
$\Rightarrow q=\frac{c}{a}$
$\Rightarrow c=qa$
Hence, we can say our assumption was wrong, so $c\ne qa$
Also, from eqn (1) and (3), we get
$\alpha +1=-2p$ and $\alpha +1=\frac{-2b}{a}$
$\Rightarrow p=\frac{b}{a}$
$\Rightarrow b=ap$
Hence, again our assumption was wrong , so $b\ne ap$
Hence, Statement 2 is true but is not the correct explanation of Statement 1 .