The condition that one root of ax2 + bx +C =0 may be triple of the other is
1) b^2 = Зас
2) 2b^2 = 9ас
3) 2b^2 = Зас
4) Зb^2 = 1 бас
pls give correct answer with solutions
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=> alpha+3alpha = -b/a
=> 4alpha = -b/a
=> alpha = -b/4a ------(1)
=> alpha*3alpha = c/a
=> 3(alpha)^2 = c/a -----(2)
substitute (1) in (2)
=> 3(-b/4a)^2 = c/a
=> b^2/16a^2 = c/3a
=> b^2 = (16a^2*c)/3a
=> b^2 = 16ac/3
=> 3b^2 = 16ac (hence proved)
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