Question

If $a, b, c$ are positive rational numbers such that $a>\,b>\,c$ and the quadratic equation
$(a+b-2 c) x^{2}+(b+c-2 a) x+(c+a-2 b)=0$ has a root in the interval $(-1,0)$, then which of the following is not correct?

• A. $c+a<\,2b$
• B. Both roots of the given equation are rational
• C. The equation $a x^{2}+2 b x+c=0$ has both negative real roots
• D. None of these

Answer 1

Answer: (D)

$a>\,b>\,c\, \dots(i)$
and given equation is
$(a+b-2 c) x^{2}+(b+c-2 a) x+(c+a-2 b)=0 \, \dots(ii)$
$\because$ Equation (ii) has a root in the interval $(-1,0)$
$\therefore f(-1) f(0)<\,0$
$ \Rightarrow (2 a-b-c)(c+a-2 b)<\,0 \, \dots(iii)$
From (i), $a>\,b$
$ \Rightarrow a-b>\,0$ and
$a>\,c \Rightarrow a-c>\,0$
$\therefore 2 a-b-c>\,0\, \dots(iv)$
From (iii) and (iv), $c+a-2 b<\,0$ or $c+a<\,2 b$. Option (a) is correct. Again, the sum of coefficients of the equation =0, that is one root is land the other root is $\frac{c+a-2 b}{a+b-2 c}$, which is a rational number as $a, b, c$ are rational. Hence, both the roots of the equation are rational
$\Rightarrow (b)$ is correct. Further, the discriminant of equation
$a x^{2}+2 b x+c=0$ is $D=4 b^{2}-4 a c$
As deduced earlier, $c+a<\,2 b$
$\Rightarrow 4 b^{2}>\,(c+a)^{2} $
$\Rightarrow 4 b^{2}>\,c^{2}+a^{2}+2 a c$
$\Rightarrow 4 b^{2}-4 a c>\,c^{2}+a^{2}-2 a c=(c-a)^{2}$
$\Rightarrow 4 b^{2}-4 a c>\,0$
$ \Rightarrow D>\,0$
Also, each of $a, b, c$ are positive.
$\therefore $ The equation $a x^{2}+2 b x+c=0$ has real and negative roots. So (c) is also correct.