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प्रश्न
Find the values of k for which the roots are real and equal in each of the following equation:
(2k + 1)x2 + 2(k + 3)x + (k + 5) = 0
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उत्तर
The given quadric equation is (2k + 1)x2 + 2(k + 3)x + k + 5 = 0, and roots are real and equal
Then find the value of k.
Here,
a = (2k + 1), b = 2(k + 3) and c = k + 5
As we know that D = b2 - 4ac
Putting the value of a = (2k + 1), b = 2(k + 3) and c = k + 5
={2(k + 3)}2 - 4 x (2k + 1) x (k + 5)
= {4(k2 + 6k + 9)} - 4(2k2 + 11k + 5)
= 4k2 + 24k + 36 - 8k2 - 44k - 20
= -4k2 - 20k + 16
The given equation will have real and equal roots, if D = 0
-4k2 - 20k + 16 = 0
-4(k2 + 5k - 4) = 0
k2 + 5k - 4 = 0
Now factorizing the above equation
k2 + 5k - 4
`k=(-b+-sqrt(b^2-4ac))/(2a)`
`k=(-5+-sqrt(25+16))/2`
`k=-5+-sqrt41/2`
So, either
Therefore, the value of `k=-5+-sqrt41/2`
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