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

• A. Both the roots of the given equation are rational
• B. $c+a<2b$
• C. The equaiton $a{x}^2+2bx+c=0$ has both negative real roots
• D. $b+c>a$

Answer 1

Answer: (A), (B), (C)

The equaiton $a{x}^2+2bx+c=0$ has both negative real roots

$a>b>c...\left(i\right)$
and given equation is
$(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0...\left(ii\right)$
Now, here the sum of coefficients
$a+b-2c+b+c-2a+c+a-2a=0$
Hence, $1$ is one of the roots of the equation.
Thus, other root is given by $\frac{c+a-2b}{a+b-2c}$
$\Rightarrow -1<\frac{c+a-2b}{a+b-2c}<0$
Also, $a>b>c\Rightarrow a+b-2c>0$
Thus, $-1<c+a-2b<0$
Or $c+a<2b$
Now, for roots of $a{x}^2+2bx+c=0$
Sum of roots $=-\frac{2b}{a}<0$
Then $\alpha +\beta =-\frac{2b}{a}<0\therefore \alpha +\beta <0\&\alpha \beta =\frac{c}{a}>0(\because a>b>c)$
and discriminant of
$a{x}^2+2bx+c=0is4b^2-4ac>0(\because b>candb<a)$
Hence, roots of $a{x}^2+2bx+c=0$ are negative and real.
Alternative (d) is correct