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).
$\Rightarrow 2(c+a)+b+c=0$
$\Rightarrow 2(2b)+b+c=0\Rightarrow c=-5b[\because a,b,c\text{ are in }AP]$
Also, $a+c=2b$
$\Rightarrow a=2b-c=2b+5b=7b$
$\therefore a^2=49b^2,c^2=25b^2$
This, show that $a^2,c^2,b^2$ are in A.P.