Complex Numbers And Quadratic Equations question 729
Question: The roots of the given equation $ (p-q)x^{2}+(q-r)x+(r-p)=0 $ are [RPET 1986; MP PET 1999; Pb. CET 2004]
Options:
A) $ \frac{p-q}{r-p},1 $
B) $ \frac{q-r}{p-q},1 $
C) $ \frac{r-p}{p-q},1 $
D) $ 1,\frac{q-r}{p-q} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation is $ (p-q)x^{2}+(q-r)x+(r-p)=0 $ $ x=\frac{(r-q)\pm \sqrt{{{(q-r)}^{2}}-4(r-p)(p-q)}}{2(p-q)} $
Þ $ x=\frac{(r-q)\pm (q+r-2p)}{2(p-q)}\Rightarrow x=\frac{r-p}{p-q},1 $
BETA
