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

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(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

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(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

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