Question

If $a,b,c$ are positive rational numbers such that $a>b>c$ and the quadratic equation
$(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ 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 $a{x}^2+2bx+c=0$ has both negative real roots
• D. None of these

Answer 1

Answer: (D)

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