# Example 8.36: Quadratic equation with real roots

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We often simulate data in SAS or R to confirm analytical results. For example, consider the following problem from the excellent text by Rice:**SAS and R**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Let U1, U2, and U3 be independent random variables uniform on [0, 1]. What is the probability that the roots of the quadratic U1*x^2 + U2*x + U3 are real?

Recall that for a quadratic equation A*x^2 + B*x + C to have real roots we need the discriminant (B^2-4AC) to be non-negative.

The answer given in the second and third editions of Rice is 1/9. Here’s how you might get there:

Since, B^2 > 4*A*C <=> B > 2*sqrt(A*C), we need to integrate B over the range 2*sqrt(a*c) to 1, then integrate over all possible values for A and C (each from 0 to 1).

Another answer can be found by taking y = b^2 and w = 4ac and integrating over their joint distribution (they’re independent, of course). That leads to an answer of approximately 0.254. Here’s how to calculate this in R:

f = function(x) { A = x[1]; B = x[2]; C = x[3]; return(as.numeric(B^2 > 4*A*C)) } library(cubature) adaptIntegrate(f, c(0,0,0), c(1,1,1), tol=0.0001, max=1000000)

which generates the following output:

$integral [1] 0.2543692 $error [1] 0.005612558 $functionEvaluations [1] 999999 $returnCode [1] -1

We leave the details of the calculations aside for now, but both seem equally plausible, at first glance. A quick simulation can suggest which is correct.

For those who want more details, here’s a more complete summary of this problem and solution.

**SAS**

Neither the SAS nor the R code is especially challenging.

data test; do trial = 1 to 10000; u1 = uniform(0); u2 = uniform(0); u3 = uniform(0); res = u2**2 - 4*u1*u3; realroot = (res ge 0); output; end; run; proc print data=test (obs=10); run; proc means data=test; var realroot; run;

Leading to the result:

The MEANS Procedure Analysis Variable : realroot N Mean Std Dev Minimum Maximum ----------------------------------------------------------------- 10000 0.2556000 0.4362197 0 1.0000000 -----------------------------------------------------------------

**R**

numsim = 10000 u1 = runif(numsim); u2 = runif(numsim); u3 = runif(numsim) res = u2^2 - 4*u1*u3 realroots = res>=0 table(realroots)/numsim

With the result

realroots FALSE TRUE 0.747 0.253

The simulation demonstrates both that the first solution is incorrect. Here the simulation serves as a valuable check for complicated analysis.

Insights into where the 1/9 solution fails would be welcomed in the comments.

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