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प्रश्न
Find the values of k for which the roots are real and equal in each of the following equation:
(k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0
उत्तर
The given quadric equation is (k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0, and roots are real and equal
Then find the value of k.
Here,
a = k + 1, b = -2(3k + 1)x and c = 8k + 1
As we know that D = b2 - 4ac
Putting the value of a = k + 1, b = -2(3k + 1)x and c = 8k + 1
= (-2(3k + 1))2 - 4 x (k + 1) x (8k + 1)
= 4(9k2 + 6k + 1) - 4(8k2 + 9k + 1)
= 36k2 + 24k + 4 - 32k2 - 36k - 4
= 4k2 - 12k
The given equation will have real and equal roots, if D = 0
4k2 - 12k = 0
k2 - 3k = 0
Now factorizing of the above equation
k(k - 3) = 0
So, either
k = 0
Or
k - 3 = 0
k = 3
Therefore, the value of k = 0, 3.
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