Question

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​

Answer 1

=> 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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