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Q.

If one root of x2+Kx + 27=0may be triple the other, then K =

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a

±  6

b

±  8

c

±  12

d

±  10

answer is C.

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Detailed Solution

Given m:n = 1:3 where m,n are roots of the equation.We have (m+n)2mn=b2ac⇒163=K227⇒K2=144⇒K=±12
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