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प्रश्न
Find the values of k for which the roots are real and equal in each of the following equation:
x2 - 2kx + 7k - 12 = 0
उत्तर
The given quadric equation is x2 - 2kx + 7k - 12 = 0, and roots are real and equal
Then find the value of k.
Here,
a = 1, b = -2k and c = 7k - 12
As we know that D = b2 - 4ac
Putting the value of a = 1, b = -2k and c = 7k - 12
= (-2k)2 - 4 x (1) x (7k - 12)
= 4k2 - 28k + 48
The given equation will have real and equal roots, if D = 0
4k2 - 28k + 48 = 0
4(k2 - 7k + 12) = 0
k2 - 7k + 12 = 0
Now factorizing of the above equation
k2 - 7k + 12 = 0
k2 - 4k - 3k + 12 = 0
k(k - 4) - 3(k - 4) = 0
(k - 3)(k - 4) = 0
So, either
k - 3 = 0
k = 3
Or
k - 4 = 0
k = 4
Therefore, the value of k = 4, 3.
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