In the following second order equation $ax^2+2bx+1.5=0$ where $a$ and $b$ are given by random points $(a,b)$ in the $[0,2]times[0,1]$ rectangle, what is the probability of having two real solutions?

I'm a little lost here. I tried integrating $4b^2-6a$ with $a=0to 2$ and $b=0to 1$ as limits but the integral comes up negative.

I created a simulation of the problem using matlab and the probability is 0.11 but I want to find a way to solve it on paper and not with using matlab.

Any thoughts?

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#### Best Answer

The quadratic formula tells us this equation has two real solutions exactly when the discriminant $4b^2 – 6a$ is positive. This describes a set of points $R$ in the $(a,b)$ plane, shown in blue here:

When the joint density of $(a,b)$ is $f$, then (by definition of pdf) the probability of any event (like $R$) is given by its integral $int_R f(a,b)da db$. Because the distribution is uniform, this is the same as finding the area of the shaded region as a proportion of the total area (equal to $2$), often written as a double integral like

$$frac{1}{2}int_0^1 int_{alt 4b^2/6, 0le ale 2} da db.$$

However, we can reduce this to a single integral: the shaded region is bounded by the parabola $a = 2 b^2/3$ on the right, $a=0$ on the left, and extends from $b=0$ to $b=1$. Its area therefore is $int_0^1 2b^2/3 db = 2/9$. That amounts to $1/9 = 0.1111ldots$ of the total area.

**Edit** (*see comments*). In case this is unclear, we can proceed more formally. The uniform distribution function $f$ is obtained by knowing (a) it is constant on the rectangle $[0,2]times[0,1]$ and (b) is zero outside this set. From (a) and the fact that any PDF must integrate to unity forces $f(a,b)=1/2$ inside the rectangle, whence

$$eqalign{ f(a,b) = &1/2, &0 le a le 2, 0 le b le 1\ &0 &text{otherwise}. }$$

Integration is defined in terms of characteristic functions: the integral over an event $E$ with respect to a measure $dmu$, written, $int cdots int_E dmu$, equals $int cdots int I_E(x) dmu(x)$ where the multiple integral is taken over all possible values of $x$ and $I_E(x) = 1$ when $xin E$ and $I_E(x)=0$ otherwise. The figure *immediately* shows that the solution is

$${Pr}_f[(a,b)in R] = int int_R f(a,b) da db$$

and the original double integral expression follows *immediately* from this expression by the definition of integration. For more about this, consult any textbook on measure theory and integration or–for a less formal approach–consult any advanced calculus text that covers multiple integration. **End** of edit.

This is essentially problem #50 from Fred Mosteller's *Fifty Challenging Problems in Probability*:

What is the probability that the quadratic equation $x^2 + 2bx + c = 0$ has real roots?

Solving it requires proposing some "reasonable" probability distribution for $(b,c)$. Mosteller chooses a set of uniform distributions over a sequence of rectangles that grows without bound and takes the limit.

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