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

The roots of the equation $x^{2}+px+q=0$ are equal if:

If $8, 2$ are the roots of ${x^2} + ax + \beta = 0$ and $3, 3$ are the roots of ${x^2} + \alpha x + b = 0$,then the roots of ${x^2} + ax + b = 0$ are

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

The number of solutions of the equation $2^x = x^2$ is

Solve the given two equations and select the correct option.
$I.$ $x = \frac{\sqrt{256}}{\sqrt{576}}$
$II.$ $3y^2 + y - 2 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module