Complex Numbers And Quadratic Equations question 25
Question: Let $ \alpha ,\beta $ be the roots of $ x^{2}+(3-\lambda )x-\lambda =0. $ The value of $ \lambda $ for which $ {{\alpha }^{2}}+{{\beta }^{2}} $ is minimum, is [AMU 2002]
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: C
Solution:
$ \alpha +\beta =\lambda -3\text{and }\alpha \beta =-\lambda $ $ {{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $ = $ {{(\lambda -3)}^{2}}+2\lambda $ = $ {{\lambda }^{2}}-4\lambda +9 $ from options, for $ \lambda =0,{{({{\alpha }^{2}}+{{\beta }^{2}})}{\lambda =0}}=9 $ for $ \lambda =1,{{({{\alpha }^{2}}+{{\beta }^{2}})}{\lambda =1}}=1-4+9=6 $ for $ \lambda =2,{{({{\alpha }^{2}}+{{\beta }^{2}})}{\lambda =2}}=4-8+9=5 $ for $ \lambda =3,{{({{\alpha }^{2}}+{{\beta }^{2}})}{\lambda =3}}=9-12+9=6 $ $ {{\alpha }^{2}}+{{\beta }^{2}} $ is minimum for $ \lambda =2 $ .
BETA
