A) \[\frac{1}{2}({{a}^{2}}+{{b}^{2}})\]
B) \[-\frac{1}{2}({{a}^{2}}+{{b}^{2}})\]
C) \[\frac{1}{2}ab\]
D) \[-\frac{1}{2}ab\]
Correct Answer: B
Solution :
The equation is \[{{x}^{2}}+x(a+b-2c)+ab-ac-bc=0\] Let its roots be\[\alpha ,\,\,\beta ,\]then \[\alpha +\beta =0\] (given) \[\Rightarrow \] \[c=\frac{a+b}{2}\] ? (i) Now, \[\alpha \beta =ab-ac-bc=ab-c(a+b)\] \[=-\frac{1}{2}({{a}^{2}}+{{b}^{2}})\] [using Eq. (i)]
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