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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:
7x2 + kx -3 = 0; x = `(2)/(3)`
Solution
7x2 + kx -3 = 0; x = `(2)/(3)`
Putting x = `(2)/(3)` in L.H.S. of equation
L.H.S. = `7 xx (2/3)^2 + (2)/(3)"k" - 3` = 0
⇒ `(28)/(9) + (2)/(3)"k" - 3` = 0
⇒ `(28 + 6"k" - 27)/(9)` = 0
⇒ 6k + 1 = 0
Hence, k = `-(1)/(6)`
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