Question

If $a, b, c$ are in A.P. and if the equations
$(b-c) x^2+(c-a) x+(a-b)=0 $ (1)
and $2(c+a) x^2+(b+c) x=0$(2)
have a common root, then

• A. $a^2, b^2, c^2$ are in A.P.
• B. $a^2, c^2, b^2$ are in A.P.
• C. $c^2, a^2, b^2$ are in A.P.
• D. none of these

Answer 1

Answer: (B)

Clearly $x=1$ is a root of (1). If $\alpha$ is the other root of (1), then
$\alpha =1 \cdot \alpha=\frac{a-b}{b-c} \text { [product of roots] } $
$ =1[\because a, b, c \text { are in A. P.] }$
Thus, the roots of (1) are 1,1 .
Now, (1) and (2) will have a common root if 1 is also a root of (2).
$\Rightarrow 2(c+a)+b+c=0 $
$\Rightarrow 2(2 b)+b+c=0 \Rightarrow c=-5 b [\because a, b, c \text { are in } AP ]$
Also, $ a+c=2 b$
$\Rightarrow a=2 b-c=2 b+5 b=7 b $
$\therefore a^2=49 b^2, c^2=25 b^2$
This, show that $a^2, c^2, b^2$ are in A.P.