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

Answer 1

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

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

$a>b>c...\left(i\right)$
and given equation is
$(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0...\left(ii\right)$
$\because$ Eq. (ii) has a root in the interval (-1, 0)
$\therefore$ f (-1) f(0) < 0
$\Rightarrow (2a-b-c)(c+a-2b)<0...\left(iii\right)$
From Eq. (i)
$a>b\Rightarrow a-b>0$
and $a>c\Rightarrow a-c>0\therefore 2a-b-c>0....\left(iv\right)$
From Eqs. (iii) and (iv),
$c+a-2b<0$
or $c+a<2b$. $\therefore$ Alternate (b) is correct
Discriminant of Eq. (ii)
$=(b+c-2a{)}^2-4(a+b-2c\left)\right(c+a-2b)=\{(b-a)+(c-a){\}}^2-4\left\{\right(a-c)+(b-c\left)\right\}\{c-b+a-b\}=\left\{\right(c-a)-(a-b){\}}^2-4\{(b-c)-(c-a)\}\times \{(a-b)-(b-c)\}Ifb-c=X,c-a=Yanda-b=Z$
then, $discriminant=(Y-Z{)}^2-4(X-Y\left)\right(Z-X)=(2X-Y-Z{)}^2(\because X+Y+Z=0)=(0+3X{)}^2=9X^2=9(b-c{)}^2$
$\therefore$ Both roots of the given equation are rational.
Alternate (c) is correct.
If roots of $a{x}^2+2bx+c=0$ are $\alpha and\beta$
Then $\alpha +\beta =-\frac{2b}{a}<0\therefore \alpha +\beta <0and\alpha \beta =\frac{c}{a}>0(\because a>b>c)$
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