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

(B) For the quadratic equation $(k-2)x^2 + 8x + k+4 = 0$ to have real and distinct roots,the discriminant $D > 0$:$D = 8^2 - 4(k-2)(k+4) > 0$$64 - 4(k

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) For the quadratic equation $(k-2)x^2 + 8x + k+4 = 0$ to have real and distinct roots,the discriminant $D > 0$:$D = 8^2 - 4(k-2)(k+4) > 0$$64 - 4(k^2 + 2k - 8) > 0$$16 - (k^2 + 2k - 8) > 0$$-k^2 - 2k + 24 > 0 \Rightarrow k^2 + 2k - 24 $(k+6)(k-4) For the roots to be negative,the sum of roots $\alpha + \beta = -\frac{8}{k-2} 0$.From $\alpha + \beta 0 \Rightarrow k > 2$.From $\alpha \beta > 0$: $\frac{k+4}{k-2} > 0 \Rightarrow k 2$.Combining all conditions: $(-6 2)$ $AND$ $(k 2)$.The intersection is $2 Among the given options,only $k = 3$ satisfies $2 < k < 4$.

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 number of distinct quadratic equations $ax^2 + bx + c = 0$ with unequal real roots that can be formed by choosing the coefficients $a, b, c$ such that $a \neq b \neq c$ from the set $\{0, 1, 2, 4\}$ is:

If $(x-2)$ is a common factor of the expressions $x^2+ax+b$ and $x^2+cx+d$,then $\frac{b-d}{c-a}$ is equal to

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

Let $a, b, c \in \mathbb{R}$ and $\alpha, \beta$ be the real roots of the equation $ax^2 + bx + c = 0$. If $a < 0, b > 0, c > 0$ and $\alpha < \beta$,then:

The quadratic equation $2x^{2} - (a^{3} + 8a - 1)x + a^{2} - 4a = 0$ has roots of opposite signs. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module