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Question: The numbers $B$ and $C$ are chosen at random between $-1$ and $1$, independently of each other. What is the probability that the quadratic equation $$x^2 + Bx + C = 0$$ has real roots? Also, derive a general expression for this probability when B and C are chosen at random from the interval $(-q, q)$ for and $q>0$.

My approach: since we're trying to find the probability of real roots. We should first realize when it has imaginary roots. So, $$B^2 - 4aC < 0$$ $$a = 1$$ $$B^2 < 4C $$ I'm not sure where to go from here. How do I now find the probability of $B$ being greater than $4C$?

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marked as duplicate by grand_chat, Lord Shark the Unknown, user91500, Adrian Keister, Micah Sep 1 '18 at 16:08

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    $\begingroup$ Look at the square in the $(B,C)$ plane with vertices $(\pm1,\pm1)$, and work out the area of the pieces the curve $B^2=4C$ divides it into. $\endgroup$ – Lord Shark the Unknown Aug 31 '18 at 15:40
  • $\begingroup$ By calculating the area defined by this inequality on the $(B,C)$-plane. $\endgroup$ – MigMit Aug 31 '18 at 15:40
  • $\begingroup$ @user170231 In that problem all three coefficients become uniform, so after we divide out the leading coefficient those that remain have ratio distributions very different from the ones used in this problem. $\endgroup$ – J.G. Aug 31 '18 at 17:30
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I assume you mean that $B,\,C\sim U(-1,\,1)$. We'll get the answer as a function of a fixed value for $C$, then average it out. For $C<0$ (which has probability $1/2$), the result is $0$; for $C> 1/4$ (which has probability $3/8$), the result is $1$; for $0\le C\le\frac{1}{4}$ (which has probability $\frac{1}{8}$), the condition $-2\sqrt{C}\le B\le 2\sqrt{C}$ has probability $4\sqrt{C}$. So the final result is $$\frac{3}{8}+\frac{1}{8}\int_0^{1/4} 4\sqrt{C}dC=\frac{3}{8}+\frac{1}{3}\bigg(\frac{1}{4}\bigg)^{3/2}=\frac{5}{12}.$$

Edit: the above is the probability of non-real complex roots; the probability of real roots is $1-\frac{5}{12}=\frac{7}{12}$.

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