Complex Numbers And Quadratic Equations question 621
Question: $ 2x^{2}-(p+1)x+(p-1)=0 $ . If $ \alpha -\beta =\alpha \beta $ , then what is the value of p [Orissa JEE 2005]
Options:
A) 1
B) 2
C) 3
D) - 2
Show Answer
Answer:
Correct Answer: B
Solution:
$ 2x^{2}-(p+1)x+(p-1)=0 $ Given $ \alpha -\beta $ = $ \alpha \beta $
Þ $ {{(\alpha +\beta )}^{2}}-4\alpha \beta $ = $ {{\alpha }^{2}}{{\beta }^{2}} $
Þ $ \frac{{{(p-1)}^{2}}}{4}=\frac{{{(p+1)}^{2}}}{4}-\frac{4(p-1)}{2} $
Þ $ 2(p-1)=p\Rightarrow p=2 $ .
BETA
