Question

If $a,b,c$ are rational number $(a>b>c>0)$ and 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 statement(s) is/are correct?

• A. $a+c<2b$
• B. Both roots are rational
• C. $a{x}^2+2bx+c=0$ has both roots negative
• D. $c{x}^2+2bx+a=0$ has both roots negative

Answer 1

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

The correct options are
A $a+c<2b$
B Both roots are rational
C $a{x}^2+2bx+c=0$ has both roots negative
D $c{x}^2+2bx+a=0$ has both roots negative

The given equation has one root as 1.
$1.\alpha =\frac{c+a-2b}{a+b-2c}$
So, the other root is $x=\frac{c+a-2b}{a+b-2c}$
As $\frac{c+a-2b}{a+b-2c}<0\Rightarrow a+c<2b$
For equation $a{x}^2+2bx+c=0$
$f\left(0\right)=c>0,\frac{-2b}{a}<0$
So both roots negative
Similarly option (D) is true