Complex Numbers And Quadratic Equations question 189
Question: For what value of $ \lambda $ the sum of the squares of the roots of $ x^{2}+(2+\lambda )x-\frac{1}{2}(1+\lambda )=0 $ is minimum [AMU 1999]
Options:
A) 3/2
B) 1
C) 1/2
D) 11/4
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation is $ x^{2}+(2+\lambda )x-\frac{1}{2}(1+\lambda )=0 $ So, $ \alpha +\beta =-(2+\lambda )=0 $ and $ \alpha \beta =-( \frac{1+\lambda }{2} ) $ Now, $ {{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $
$ \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{[ -(2+\lambda ) ]}^{2}}+2.\frac{(1+\lambda )}{2} $
Þ $ {{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+4+4\lambda +1+\lambda ={{\lambda }^{2}}+5\lambda +5 $ which is minimum for $ \lambda =1/2 $ .
BETA
