Questions from http://www.sgforums.com/forums/2297/topics/312370
Show that the solutions of the equation x2 + kx = 3 - k are real for all real values of k.
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Answer:
Shift all to one side
x2 + kx - 3 + k = 0
x2 + kx + (k - 3) = 0
For all solutions to be real, we must show that the discriminant is ≥ 0
Discriminant = b2 - 4ac
= k2 - 4(1)(k - 3)
= k2 - 4k + 12
= k2 - 4k + 4 + 8
= (k - 2)2 + 8
Since (k - 2)2 ≥ 0, (k - 2)2 + 8 ≥ 0
Hence discriminant is ≥ 0
Thus, the solutions of the equation x2 + kx = 3 - k are real for all real values of k (shown).
