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mik1a
firstly, b^2-4ac > 0 for two distinct roots.
second, you shouldn't assume it is true and follow through by manipulating an equation. as a proof, you should take b^2-4ac, substitute in, rearrange it and show that it is positive irregardless of k. writing it as an equation leads you to do things like multiply by -1, which is silly because now you have 4ac-b^2, not what you want.

I do agree that the question looks like it was writen wrongly.

see above post. I misread the question.
Reply 21
mik1a
firstly, b^2-4ac > 0 for two distinct roots.
second, you shouldn't assume it is true and follow through by manipulating an equation. as a proof, you should take b^2-4ac, substitute in, rearrange it and show that it is positive irregardless of k. writing it as an equation leads you to do things like multiply by -1, which is silly because now you have 4ac-b^2, not what you want.

I do agree that the question looks like it was writen wrongly.

Even if he did prove b^2 -4ac > 0 for all values of k, but ..with k = 0. there's only 1 root, not 2 distinct roots. Hmm, u guys too mechanical,
Reply 22
Widowmaker
it is a quadratic function.
see the k^2.


hehe yes I know, but you said always use a graph for inequalities, just wires crossed then :biggrin:

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