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Question
Find the values of k for which the roots are real and equal in each of the following equation:
(k + 1)x2 + 2(k + 3)x + (k + 8) = 0
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Solution
The given quadric equation is (k + 1)x2 + 2(k + 3)x + (k + 8) = 0, and roots are real and equal
Then find the value of k.
Here,
a = (k + 1), b = 2(k + 3) and c = (k + 8)
As we know that D = b2 - 4ac
Putting the value of a = (k + 1), b = 2(k + 3) and c = (k + 8)
= (2(k + 3))2 - 4 x (k + 1) x (k + 8)
= (4k2 + 24k + 36) - 4(k2 + 9k + 8)
= 4k2 + 24k + 36 - 4k2 - 36k - 32
= -12k + 4
The given equation will have real and equal roots, if D = 0
-12k + 4 = 0
12k = 4
k = 4/12
k = 1/3
Therefore, the value of k = 1/3.
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