# Probability of second order equation with real coefficients to have real solutions?

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 $0\leq a \leq2$ and $0\leq b\leq 1$ 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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Working with the discriminant $4b^2 -6a$ is the correct idea. However you're not interested in calculating its integral over the given rectangle, but in determining where the discriminant is non-negative (more precisely, the area of the region where it's non-negative). Start by solving $4b^2 -6a \geq 0$; once you've identified the subregion of $[0,2]\times [0,1]$ where the inequality is true, you can calculate its area by an appropriate integration.

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Just to continue @lazyhaze's line of thought:

Solving for $a=\frac23b^2$, we get probability $p=\frac12\int_0^1\frac23b^2db=$ (spoiler:)

$=\frac19$

where the fraction $\frac12$ is our "normalization" constant, the reciprocal of the total area $A=(a_1-a_0)(b_1-b_0)=(2-0)(1-0)=2\cdot1=2$ over $(a,b)\in[0,2]\times[0,1]$.

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could you maybe explain the 1/2 in front of the integral ? – system Apr 1 '12 at 21:32