The value of k for which the equation (k - 2) | vedclass

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

Question

(C) For the quadratic equation $ax^2 + bx + c = 0$ to have real and distinct roots,the discriminant $D = b^2 - 4ac > 0$.Here,$a = k - 2$,$b = 8$,and $

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For the quadratic equation $ax^2 + bx + c = 0$ to have real and distinct roots,the discriminant $D = b^2 - 4ac > 0$.Here,$a = k - 2$,$b = 8$,and $c = k + 4$.$D = 8^2 - 4(k - 2)(k + 4) = 64 - 4(k^2 + 2k - 8) = 64 - 4k^2 - 8k + 32 = -4k^2 - 8k + 96$.Setting $D > 0$: $-4(k^2 + 2k - 24) > 0$ $\Rightarrow k^2 + 2k - 24 For roots to be negative,the sum of roots $\alpha + \beta = -b/a = -8/(k - 2)  0$.From $-8/(k - 2)  0 \Rightarrow k > 2$.From $(k + 4)/(k - 2) > 0$,we get $k  2$.Combining all conditions: $2 Checking $k = 3$: $(3 - 2)x^2 + 8x + (3 + 4) = 0$ $\Rightarrow x^2 + 8x + 7 = 0$ $\Rightarrow (x + 1)(x + 7) = 0$,giving $x = -1, -7$,which are real,distinct,and negative.

The value of $k$ for which the equation $(k - 2)x^2 + 8x + k + 4 = 0$ has both roots real,distinct,and negative is

Similar Questions

The sum of the roots of the equation $x^2 + |2x - 3| - 4 = 0$ is:

The positive integer $n$, for which the solutions of the equation $x(x+2)+(x+2)(x+4)+...+(x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is:

The number of real roots of $|x|^2-5|x|+6=0$ is

The roots of the equation $x^2 + 2\sqrt{3}x + 3 = 0$ are

Let $a \ne a_1 \ne 0,$ $f(x) = ax^2 + bx + c,$ $g(x) = a_1x^2 + b_1x + c_1,$ and $p(x) = f(x) - g(x).$ If $p(x) = 0$ only for $x = -1$ and $p(-2) = 2,$ then the value of $p(2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module