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प्रश्न
Find the values of k for which the roots are real and equal in each of the following equation:
(3k+1)x2 + 2(k + 1)x + k = 0
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उत्तर
The given quadric equation is (3k+1)x2 + 2(k + 1)x + k = 0, and roots are real and equal
Then find the value of k.
Here, a = (3k + 1), b = 2(k + 1) and c = k
As we know that D = b2 - 4ac
Putting the value of a = (3k + 1), b = 2(k + 1) and c = k
= (2(k + 1))2 - 4 x (3k + 1) x (k)
= 4(k2 + 2k + 1) - 4k(3k + 1)
= 4k2 + 8k + 4 - 12k2 - 4k
= -8k2 + 4k + 4
The given equation will have real and equal roots, if D = 0
Thus,
-8k2 + 4k + 4 = 0
-4(2k2 - k - 1) = 0
2k2 - k - 1 = 0
Now factorizing of the above equation
2k2 - k - 1 = 0
2k2 - 2k + k - 1 = 0
2k(k - 1) + 1(k - 1) = 0
(k - 1)(2k + 1) = 0
So, either
k - 1 = 0
k = 1
Or
2k + 1 = 0
2k = -1
k = -1/2
Therefore, the value of k = 1, -1/2
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