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$2x^3+3x^2-x+5-m=0$

I know for the above equation there is the following condition for the case when all the three roots must be distinct and real:

$D = -4b^3d + b^2c^2 - 4 ac^3 + 18abcd - 27a^2d^2 > 0 $

So, we calculate $D$ and then we find out $m$.

But is there another way to solve this problem? Maybe with the Vieta's formulas?

Florin M.
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1 Answers1

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You have three real solutions if and only if the curve has a positive maximum and a negative minimum (draw a picture). Can you do it now?

preferred_anon
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  • I was wondering if it's possible without drawing the function graph – Florin M. Jun 25 '15 at 12:19
  • Drawing the graph is a very good idea, and I don't know why you wouldn't want to do it. You can prove the statement above without drawing the graph by using Rolle's Theorem and the Intermediate Value Theorem. Why don't you want to draw the graph? – preferred_anon Jun 25 '15 at 12:30