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प्रश्न
In the following, find the value of k for which the given value is a solution of the given equation:
`kx^2+sqrt2x-4=0`, `x=sqrt2`
उत्तर
We are given here that,
`kx^2+sqrt2x-4=0`, `x=sqrt2`
Now, as we know that `x=sqrt2` is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting `x=sqrt2` in the above equation gives us,
`kx^2+sqrt2x-4=0`
`k(sqrt2)^2+sqrt2(sqrt2)-4=0`
2k + 2 - 4 = 0
2k - 2 = 0
2k = 2
`k=2/2`
k = 1
Hence, the value of k = 1.
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