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Question
Find the values of k for which the roots are real and equal in each of the following equation:
4x2 - 2(k + 1)x + (k + 4) = 0
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Solution
The given quadric equation is 4x2 - 2(k + 1)x + (k + 4) = 0, and roots are real and equal
Then find the value of k.
Here,
a = 4, b = -2(k + 1) and c = k + 4
As we know that D = b2 - 4ac
Putting the value of a = 4, b = -2(k + 1) and c = k + 4
= {-2(k + 1)}2 - 4 x 4 x (k + 4)
= {4(k2 + 2k + 1)} - 16(k + 4)
= 4k2 + 8k + 4 - 16k - 64
= 4k2 - 8k - 60
The given equation will have real and equal roots, if D = 0
4k2 - 8k - 60 = 0
4(k2 - 2k - 15) = 0
k2 - 2k - 15 = 0
Now factorizing of the above equation
k2 - 2k - 15 = 0
k2 + 3k - 5k - 15 = 0
k(k + 3) - 5(k + 3) = 0
(k + 3)(k - 5) = 0
So, either
k + 3 = 0
k = -3
Or
k - 5 = 0
k = 5
Therefore, the value of k = -3, 5.
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