If k in ( - infty , - 2) cup (2, infty ), the | vedclass

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

Question

(B) The given quadratic equation is $x^2 + 2kx + 4 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the nature of the roots is determined by the disc

Vedclass · Accepted Answer

Real and unequal

Vedclass · Answer

(B) The given quadratic equation is $x^2 + 2kx + 4 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the nature of the roots is determined by the discriminant $D = b^2 - 4ac$.Here,$a = 1$,$b = 2k$,and $c = 4$.$D = (2k)^2 - 4(1)(4) = 4k^2 - 16$.For the roots to be real and unequal,we must have $D > 0$.$4k^2 - 16 > 0 \Rightarrow 4(k^2 - 4) > 0 \Rightarrow k^2 > 4$.This inequality holds when $k > 2$ or $k Thus,$k \in ( - \infty , - 2) \cup (2, \infty )$.Therefore,the roots are real and unequal.

If $k \in ( - \infty , - 2) \cup (2, \infty ),$ then the roots of the equation $x^2 + 2kx + 4 = 0$ are

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt[3]{2197}$
$II.$ $2y^2 - 54y + 364 = 0$

If the product of the roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$,then for what value of $k$ will its roots be real?

The values of $a$ and $b$ for which the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots are:

If $x^{4} + \frac{1}{x^{4}} = 119$,then the value of $x^{3} - \frac{1}{x^{3}}$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt[3]{4913}$
$II.$ $13y + 3x = 246$

Vedclass Test Series

Exam Paper Generator

Online Exam Module