Complex Numbers And Quadratic Equations question 670
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?
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 )=0and\alpha \beta =-( \frac{1+\lambda }{2} ) $ Now, $ {{\alpha }^{2}}+{{\beta }^{2}}={{( a+\beta )}^{2}}-2\alpha \beta $
$ \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{[ -(2+\lambda ) ]}^{2}}+2\frac{(1+\lambda )}{2} $
$ \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+4+4\lambda +1+\lambda $ $ ={{\lambda }^{2}}+5\lambda +5 $ Which is minimum for $ \lambda =1/2. $
BETA
