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प्रश्न
Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation
`(K + 2)x^2 - kx + 6 = 0`
उत्तर
`(K + 2)x^2 - Kx + 6 = 0` ...(1)
Substituting x = 3 in equation (1), we get
`(K + 2) (3)^2 - K(3) + 6 = 0`
`:. 9(K +2) - 3K + 6 = 0`
`:. 9K + 18 - 3k + 6 = 0`
`:. 6k + 24 = 0`
`:. K= -4`
Now, substituting K = –4 in equation (1), we get
`(-4 + 2)x^2 - (-4)x + 6 = 0`
`:. -2x^2 + 4x + 6 = 0`
`:. x^2 - 2x - 3 = 0`
`:. x^2 - 3x + x - 3 = 0`
`:. x(x - 3) + 1(x- 3) = 0`
`:. (x + 1) (x + 3) = 0`
So, the roots are x = –1 and x = 3.
Thus the other root of the equation is x = -1.
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