Complex Numbers And Quadratic Equations question 565
Question: If $ \alpha $ and $ \beta $ are the roots of the equation $ ax^{2}+bx+c=0 $ $ (a\ne 0; $ $ a,b,c $ being different), then $ (1+\alpha +{{\alpha }^{2}}) $ $ (1+\beta +{{\beta }^{2}}) $ = [DCE 2000]
Options:
A) Zero
B) Positive
C) Negative
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \alpha +\beta =-b/a,\alpha \beta =c/a $ $ (1+\alpha +{{\alpha }^{2}})(1+\beta +{{\beta }^{2}}) $ $ =1+(\alpha +\beta )+({{\alpha }^{2}}+{{\beta }^{2}})+\alpha \beta +\alpha \beta (\alpha +\beta )+{{\alpha }^{2}}{{\beta }^{2}} $ $ =1+(\alpha +\beta )+{{(\alpha +\beta )}^{2}}-\alpha \beta +\alpha \beta (\alpha +\beta )+{{(\alpha \beta )}^{2}} $ $ =1-\frac{b}{a}+\frac{b^{2}}{a^{2}}-\frac{c}{a}+( \frac{c}{a} )( -\frac{b}{a} )+\frac{c^{2}}{a^{2}} $ $ =(a^{2}+b^{2}+c^{2}-ab-bc-ca)/a^{2} $ $ =[(a-b^{2})+{{(b-c)}^{2}}+{{(c-a)}^{2}}]/2a^{2} $ which is positive. Trick: It is almost clear that for every different values of a,b,c the function is zero, positive or negative. Therefore let $ a=1,b=0,c=-4 $ so that $ \alpha =2,\ \beta =-2 $ . Obviously the expression has positive value.
BETA
