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Question
For what value of k, (4 - k)x2 + (2k + 4)x + (8k + 1) = 0, is a perfect square.
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Solution
The given quadric equation is (4 - k)x2 + (2k + 4)x + (8k + 1) = 0, and roots are real and equal
Then find the value of k.
Here, a = (4 - k), b = (2k + 4) and c = (8k + 1)
As we know that D = b2 - 4ac
Putting the value of a = (4 - k), b = (2k + 4) and c = (8k + 1)
= (2k + 4)2 - 4 x (4 - k) x (8k + 1)
= 4k2 + 16k - 16 - 4(4 + 31k - 8k2)
= 4k2 + 16k - 16 - 16 - 124k + 32k2
= 36k2 - 108k + 0
= 36k2 - 108k
The given equation will have real and equal roots, if D = 0
Thus,
36k2 - 108k = 0
18k(2k - 6) = 0
k(2k - 6) = 0
Now factorizing of the above equation
k(2k - 6) = 0
So, either
k = 0
or
2k - 6 = 0
2k = 6
k = 6/2 = 3
Therefore, the value of k = 0, 3.
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