Question

If $a , b , c$ are positive real numbers such that $a > b > c$ and the quadratic equation $(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$ has a root in the interval $(-1,0)$ then the roots of the equation $x ^2+( a + c ) x +4 b ^2=0$ are

• A. imaginary
• B. real and unequal
• C. real and equal
• D. less than -1 .

Answer 1

Answer: (A)

Let $f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)$
As $ a>b>c \Rightarrow a+b-2 c>0$
$\Rightarrow f (-1) f (0)<0$
$\Rightarrow (2 a - b - c )( c + a -2 b )<0$ .....(1)
Also $a>b \Rightarrow a-b>0$ and $a>c \Rightarrow a-c>0$
$\therefore 2 a - b - c >0$ .....(2)
$\therefore $ From (1) and (2), we get
$c + a -2 b <0 \Rightarrow c + a <2 b$ ....(3)
Now, discriminant of given equation $=(a+c)^2-16 b^2=(a+c)^2-4 b^2-12 b^2<0$
(As $a + c <2 b$ ).