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Question
In each of the following find the values of k of which the given value is a solution of the given equation:
`"k"x^2 + sqrt(2)x- 4 = 0; x = sqrt(2)`
Solution
`"k"x^2 + sqrt(2)x- 4 = 0; x = sqrt(2)`
Putting x = `sqrt(2)` in L.H.S. of equation
L.H.S = k.`(sqrt(2))^2 + sqrt(2) xx sqrt(2) - 4` = 0
⇒ 2k + 2 - 4 = 0
⇒ 2k - 2 = 0
Hence, k = `(2)/(2)`.
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