Question

Let $a,b,c\in Q^{+}$ satisfying $a>b>c$. Which of the following statement(s) hold true for the quadratic polynomial $f\left(x\right)=(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)$?

• A. The mouth of the parabola $y=f\left(x\right)$ opens upwards.
• B. Both roots of the equation $f\left(x\right)=0$ are rational.
• C. The $x$-coordinate of vertex of the graph is positive.
• D. The product of the roots is always negative.

Answer 1

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

Both roots of the equation $f\left(x\right)=0$ are rational.
The $x$-coordinate of vertex of the graph is positive.
The mouth of the parabola $y=f\left(x\right)$ opens upwards.

We have $a,b,c$ are are positive real and $a>b>c$

And $f\left(x\right)=(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)$

As $a>b>c$we can $a-c>0$and$b-c>0$

So we get $a+b-2c>0$ which means the parabola is opening upward

Concept :-

If $g\left(x\right)=a{x}^2+bx+c$ and $a+b+c=0$ so the roots are $1,\frac{c}{a}$

In this case
also $Sum=a+b-2c+b+c-2a+c+a-2b=0$

So the roots are $1,\frac{c+a-2b}{a+b-2c}$

Both the roots are real and rational

$x$ coordinate is $2\left(\frac{2a-b-c}{a+b-2c}\right)$which is positive.