Complex Numbers And Quadratic Equations question 600
Question: If the roots of the equation $ x^{2}-5x+16=0 $ are $ \alpha ,\beta $ and the roots of equation $ x^{2}+px+q=0 $ are $ {{\alpha }^{2}}+{{\beta }^{2}}, $ $ \frac{\alpha \beta }{2}, $ then [MP PET 2001]
Options:
A) p = 1, q = - 56
B) p = - 1, q = - 56
C) p = 1, q = 56
D) p = - 1, q = 56
Show Answer
Answer:
Correct Answer: B
Solution:
Since roots of the equation $ x^{2}-5x+16=0 $ are $ \alpha ,\beta $ .
$ \Rightarrow \alpha +\beta =5 $ and $ \alpha \beta =16 $ and $ {{\alpha }^{2}}+{{\beta }^{2}}+\frac{\alpha \beta }{2}=-p $
$ \Rightarrow {{(\alpha +\beta )}^{2}}-2\alpha \beta +\frac{\alpha \beta }{2}=-p $
$ \Rightarrow 25-32+8=-p $
$ \Rightarrow p=-1 $ $ and\text{(}{{\alpha }^{2}}+{{\beta }^{2}})( \frac{\alpha \beta }{2} )=q $
$ \Rightarrow [ {{(\alpha +\beta )}^{2}}-2\alpha \beta ][ \frac{\alpha \beta }{2} ]=q $
Þ $ q=[25-32]\frac{16}{2}=-56 $ So, $ p=-1,q=-56 $ .
BETA
