Question

If $a,b,c$ are rational numbers $(a>b>c>0)$ 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 statements 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)

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

$(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0,a+b-2c\ne 0$
The given equation has one root as $1.$
Let $\alpha$ be the other root.
$1\cdot \alpha =\frac{c+a-2b}{a+b-2c}$
So, the other root is $x=\frac{c+a-2b}{a+b-2c}$
As $\frac{c+a-2b}{a+b-2c}<0$ and $a+b>2c$
$\Rightarrow c+a-2b<0$
$\Rightarrow a+c<2b$

For equation $a{x}^2+2bx+c=0,$
$f\left(0\right)=c>0,\frac{-2b}{a}<0$
So both roots are negative.

Similarly, for $c{x}^2+2bx+a=0,$ both roots are negative.