The value of k for which the quadratic equati | vedclass

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(C) The given quadratic equation is $kx^2 + 1 = kx + 3x - 11x^2$.Rearranging the terms,we get $(k + 11)x^2 - (k + 3)x + 1 = 0$.For the quadratic equat

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$5, - 7$

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(C) The given quadratic equation is $kx^2 + 1 = kx + 3x - 11x^2$.Rearranging the terms,we get $(k + 11)x^2 - (k + 3)x + 1 = 0$.For the quadratic equation to have real and equal roots,the discriminant $D$ must be equal to $0$.$D = b^2 - 4ac = 0$Substituting the coefficients $a = (k + 11)$,$b = -(k + 3)$,and $c = 1$:$(k + 3)^2 - 4(k + 11)(1) = 0$$k^2 + 6k + 9 - 4k - 44 = 0$$k^2 + 2k - 35 = 0$Factoring the quadratic equation:$(k + 7)(k - 5) = 0$Thus,$k = -7$ or $k = 5$.

The value of $k$ for which the quadratic equation $kx^2 + 1 = kx + 3x - 11x^2$ has real and equal roots is:

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