can the quadratic equation x^2+kx+k=0 have equal roots for some odd integer k>1
Answers
Answered by
5
Answer:
no ,this condition is not possible.
I hope the attached pic would help you
Attachments:
![](https://hi-static.z-dn.net/files/d17/b731c773359c87c56fc2c4624cd0b8be.jpg)
Answered by
5
Answer:No
Solution:
For Quadratic equations to be equal roots
D = 0
ie
![{b}^{2} - 4ac = 0 \\ \\ {b}^{2} - 4ac = 0 \\ \\](https://tex.z-dn.net/?f=+%7Bb%7D%5E%7B2%7D++-+4ac+%3D+0+%5C%5C++%5C%5C+)
Here in the given equation as compared to the standard Quadratic equation
![a \: = 1 \\ \\ b = k \\ \\ c \: = k \\ \\ a \: = 1 \\ \\ b = k \\ \\ c \: = k \\ \\](https://tex.z-dn.net/?f=a+%5C%3A++%3D+1+%5C%5C++%5C%5C+b+%3D+k+%5C%5C++%5C%5C+c+%5C%3A++%3D+k+%5C%5C++%5C%5C+)
So
![{k}^{2} - 4k = 0 \\ \\ k(k - 4) = 0 \\ \\ so \: either \: k = 0 \\ \\ or \\ \\ k - 4 = 0 \\ \\ k = 4 \\ \\ {k}^{2} - 4k = 0 \\ \\ k(k - 4) = 0 \\ \\ so \: either \: k = 0 \\ \\ or \\ \\ k - 4 = 0 \\ \\ k = 4 \\ \\](https://tex.z-dn.net/?f=+%7Bk%7D%5E%7B2%7D++-+4k+%3D+0+%5C%5C++%5C%5C+k%28k+-+4%29+%3D+0+%5C%5C++%5C%5C+so+%5C%3A+either+%5C%3A+k+%3D+0+%5C%5C++%5C%5C+or+%5C%5C++%5C%5C+k+-+4+%3D+0+%5C%5C++%5C%5C+k+%3D+4+%5C%5C++%5C%5C+)
So, the given Quadratic equation has real and equal roots if either k = 0 or k = 4.
So it doesn't satisfy the given condition,since 4 is not an odd number.
Hope it helps you.
Solution:
For Quadratic equations to be equal roots
D = 0
ie
Here in the given equation as compared to the standard Quadratic equation
So
So, the given Quadratic equation has real and equal roots if either k = 0 or k = 4.
So it doesn't satisfy the given condition,since 4 is not an odd number.
Hope it helps you.
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