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Question
In each of the following find the values of k of which the given value is a solution of the given equation:
x2 + 3ax + k = 0; x = a.
Solution
x2 + 3ax + k = 0; x = a.
Putting x = -a in L.H.S. of equation
L.H.S = (-a)2 + 3a x (-a) + k = 0
⇒ a2 - 3a2 + k = 0
⇒ -2a2 + k = 0
Hence, k = 2a2.
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