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Question

The equation x2+k2=3−k+2kx has normal no real solutions
Find the possible values of k

A
k>3
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B
k3
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C
k3
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D
k<3
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Solution

The correct option is A k>3
Given the equation is x2+k2=3k+2kx
or, x22kx+k2+k3=0
This equation will've no real solution if the discriminant of the equation is <0.
Then,
(2k)24(k2+k3)<0
or, k2(k2+k3)<0
or, 3k<0
or, k>3.

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