Complex Numbers And Quadratic Equations question 192
Question: If a, b be the roots of $ x^{2}+px+q=0 $ and $ \alpha +h,\beta +h $ are the roots of $ x^{2}+rx+s=0 $ , then [AMU 2001]
Options:
A) $ \frac{p}{r}=\frac{q}{s} $
B) $ 2h=[ \frac{p}{q}+\frac{r}{s} ] $
C) $ p^{2}-4q=r^{2}-4s $
D) $ pr^{2}=qs^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \alpha +\beta =-p,\alpha \beta =q $ $ \alpha +\beta +2h=-r, $ $ (\alpha +h)(\beta +h)=s $ $ -p+2h=-r\Rightarrow h=\frac{p-r}{2} $ …….(i) Now, $ \alpha \beta +h(\alpha +\beta )+h^{2}=s $
$ \Rightarrow q+h(-p)+h^{2}=s $
$ \Rightarrow q+( \frac{p-r}{2} )(-p)+{{( \frac{p-r}{2} )}^{2}}=s $
$ \Rightarrow q-\frac{(p^{2}-pr)}{2}+\frac{p^{2}+r^{2}-2pr}{4}=s $
$ \Rightarrow 4q-2p^{2}+2pr+p^{2}+r^{2}-2pr=4s $
$ \Rightarrow 4q-p^{2}+r^{2}-4s=0 $
$ \Rightarrow r^{2}-4s=p^{2}-4q $ .
BETA
