Complex Numbers And Quadratic Equations question 645
Question: The value of a for which the sum of the squares of the roots of the equation $ 2x^{2}-2(a-2)x-(a+1)=0 $ is least, is
Options:
A) 1
B) $ 3/2 $
C) 2
D) None
Show Answer
Answer:
Correct Answer: B
Solution:
If $ \alpha ,\beta $ be the roots of the equation then $ \alpha +\beta =a-2, $ $ \alpha \beta =-\frac{a+1}{2} $ Sum of square of roots $ S={{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $ $ ={{(a-2)}^{2}}+(a+1)=a^{2}-3a+5 $ $ S=a^{2}-3a+\frac{9}{4}+\frac{11}{4} $ $ S={{( a-\frac{3}{2} )}^{2}}+\frac{11}{4} $ Clearly S is least when $ a-\frac{3}{2}=0\Rightarrow a=\frac{3}{2} $
BETA
