Find the values of k for which the roots are real and equal in each of the following equations:
(4 - k)x2 + (2k + 4)x + (8k + 1) = 0
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Answered by
1
Answer:
2k answer. i hope you are help
Answered by
5
Step-by-step explanation:
Here, a = (4 - k), b = (2k + 4) and c = (8k + 1)
As we know that D = b2 - 4ac
Putting the value of a = (4 - k), b = (2k + 4) and c = (8k + 1)
= (2k + 4)2 - 4 x (4 - k) x (8k + 1)
= 4k2 + 16k - 16 - 4(4 + 31k - 8k2)
= 4k2 + 16k - 16 - 16 - 124k + 32k2
= 36k2 - 108k + 0
= 36k2 - 108k
The given equation will have real and equal roots, if D = 0
Thus,
36k2 - 108k = 0
18k(2k - 6) = 0
k(2k - 6) = 0
Now factorizing of the above equation
k(2k - 6) = 0
So, either
k = 0
or
2k - 6 = 0
2k = 6
k = 6/2 = 3
Therefore, the value of k = 0, 3.
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