SOLUTION: Find the value of a such that y^2 -6y + a is a perfect square. a=
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Question 32335
:
Find the value of a such that
y^2 -6y + a
is a perfect square.
a=
Answer by
mukhopadhyay(490)
(
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):
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y^2 -6y + a
=y^2 - 2*(3y) + a
=[y^2 - 2*y*3 + 3^2] + (a - 3^2)
=(y-3)^2 + (a-9)
For above to be a complete square, the remainder (a-9) has to be zero
Thus a-9 = 0
=> a=9