Question

If $a,b,c$ are positive real 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 the roots of the equation $x^2+(a+c)x+4b^2=0$ are

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

Answer 1

Answer: (A)

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