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show that the equation x^3-3x^2+3=0 has a solution between x=2 and x=3

Sagot :

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The equation has 3 solutions (determined by the degree of polynomial and non-extreme case).

If you use cubic equation to try and solve the equality you get

[tex]x_1\approx-0.88 \\ x_2\approx1.35 \\ x_3\approx2.53[/tex]

Clearly third one is in [tex](2,3)[/tex] so the statement that one solution is between 2 and 3 is true.

Hope this helps.