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Question Number 176449 by peter frank last updated on 19/Sep/22
A fair dice was thrown twice and  it landed on a and b respectively then  the probability that cubic equation  x^3 −(3a+1)x^2 +(3a+2b)x−2b=0  has three distinct root
$$\mathrm{A}\:\mathrm{fair}\:\mathrm{dice}\:\mathrm{was}\:\mathrm{thrown}\:\mathrm{twice}\:\mathrm{and} \\ $$$$\mathrm{it}\:\mathrm{landed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{respectively}\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{cubic}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{3}} −\left(\mathrm{3a}+\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3a}+\mathrm{2b}\right)\mathrm{x}−\mathrm{2b}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{three}\:\mathrm{distinct}\:\mathrm{root} \\ $$$$ \\ $$

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