Question

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

Answer 1

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

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

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

Now, $f\left(1\right)=a+b-2c+b+c-2a+c+a-2b$

$⟹f\left(1\right)=0$

It means that 1 is a root of the given quadratic equation.

Given that, $a>b>c$

$⟹a-c>0andb-c>0$

$⟹(a-c)+(b-c)>0$

$⟹a+b-2c>0$

We got one positive root of the given quadratic equation i.e. 1 and the other root lies between the interval (-1,0). So, the other root is negative.

$\therefore$ Product of roots should be negative

$⟹\frac{c+a-2b}{a+b-2c}<0$

$⟹c+a-2b<0$ $(\because a+b-2c>0)$

$⟹c+a<2b$

$\because a,b,c$ are rational numbers

$\therefore$ Product of roots $i.e.\frac{c+a-2b}{a+b-2c}$ is rational number and we know that one root is 1 i.e. rational. So, the other root must be a rational number.

For, $a{x}^2+2bx+c=0$

Sum of roots = $-2b/a$

Product of roots = $c/a$

$\because a,b,c$ are real

$\therefore$ Sum and product of roots are also real.

Hence, we can say that the given quadratic equation has real roots.

Now, $\because$ Sum of roots is negative and product of roots is positive.

$\therefore$ Both roots of the given quadratic equation will be negative.

Hence, option B, C, D are correct.