A) \[a(a+b)=2bc\]
B) \[c(a+c)=2ab\]
C) \[b(a+b)=2ac\]
D) \[b(a+b)=ac\]
Correct Answer: C
Solution :
Let \[\alpha \] and \[\beta \] be two roots of \[a{{x}^{2}}+bx+c=0\] Then \[\alpha +\beta =-\frac{b}{a}\]and \[\alpha \beta =\frac{c}{a}\] Þ \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta =\frac{{{b}^{2}}}{{{a}^{2}}}-2\frac{c}{a}\] So under condition \[\alpha +\beta ={{a}^{2}}+{{\beta }^{2}}\] Þ \[-\frac{b}{a}=\frac{{{b}^{2}}-2ac}{{{a}^{2}}}\]Þ \[b(a+b)=2ac\].
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