The value of k for which the quadratic equati | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$5, - 7$

Vedclass · Answer

(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:

Similar Questions

If $x$ is real,then the minimum value of $y = \frac{x^2-x+1}{x^2+x+1}$ is

The sum of the non-real roots of $(p^2+p-3)(p^2+p-2)-12=0$ is

If $\alpha, \beta$ are the irrational roots of the equation $3p^2x^3 + px^2 + qx + 3 = 0$ when $p = 1$ and $q = -7$,then $|\alpha - \beta| = $

The number of real values of $a$ satisfying the equation $a^2 - 2a\sin x + 1 = 0$ is

If the equation $x^2-3ax+a^2-2a-K=0$ has different real roots for every rational number $a$,then $K$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module