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 $D$ must be greater than $0$.$D = b^2 -

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 $D$ must be greater than $0$.$D = b^2 - 4ac = 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 product of roots $\alpha\beta = \frac{c}{a} = \frac{K + 4}{K - 2} > 0$ and the sum of roots $\alpha + \beta = -\frac{b}{a} = -\frac{8}{K - 2} From $\frac{K + 4}{K - 2} > 0$,we get $K  2$.From $-\frac{8}{K - 2}  0$,which implies $K > 2$.Combining the conditions $-6  2$,we get $2 Among the given options,only $K = 3$ satisfies this condition.

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

If $\cos^4 \theta + \alpha$ and $\sin^4 \theta + \alpha$ are the roots of the equation $x^2 + 2bx + b = 0$,and $\cos^2 \theta + \beta$ and $\sin^2 \theta + \beta$ are the roots of the equation $x^2 + 4x + 2 = 0$,then find the value of $b$.

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,the value of $\alpha^{3} + \beta^{3}$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x = 4y + 85$
$II.$ $y = \sqrt[3]{17576}$

If $\alpha, \beta$ are the roots of $(x - a)(x - b) = c,$ where $c \neq 0,$ then the roots of $(x - \alpha)(x - \beta) + c = 0$ are

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x^{2} + 11x + 14 = 0$
$II.$ $4y^{2} + 12y + 9 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module