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 a quadratic equation $ax^2 + bx + c = 0$ to have real and distinct roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$.Here,$a = (

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For a quadratic equation $ax^2 + bx + c = 0$ to have real and distinct roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$.Here,$a = (k - 2)$,$b = 8$,and $c = (k + 4)$.$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 roots to be negative,the sum of roots $-b/a = -8/(k - 2) 0$.From $-8/(k - 2) 0 \Rightarrow k > 2$.From $(k + 4)/(k - 2) > 0$,we get $k > 2$ or $k Combining these conditions with $-6 Checking the options,$k = 3$ satisfies $2 < 3 < 4$.

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

Solve the given two equations and select the correct option.

Two candidates attempt to solve the equation $x^2 + px + q = 0$. One starts with a wrong value of $p$ and finds the roots to be $2$ and $6$,and the other starts with a wrong value of $q$ and finds the roots to be $2$ and $-9$. The roots of the original equation are

Solve the given two equations and select the correct answer from the given options.
$I. \quad 4x + 7y = 209$
$II. \quad 12x - 14y = -38$

Solve the given two equations and select the correct option.
$I.$ $2x^2 + 3x + 1 = 0$
$II.$ $12y^2 + 7y + 1 = 0$

If $a + b + c = 0$,$a \neq 0$,$a, b, c \in Q$,then both the roots of the equation $ax^2 + bx + c = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module