Given ${a, b, c} \in [- L, L]$ ${a,b,c} in [-L,L]$ What is the probability that the roots of $a x^{2} + b x + c = 0$ $a x^2+b x + c = 0$ be real?

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Given ${a, b, c} \in [- L, L]$ ${a,b,c} in [-L,L]$ What is the probability that the roots of $a x^{2} + b x + c = 0$ $a x^2+b x + c = 0$ be real?

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Answer 1 · 2016-11-19T00:57:52

$\frac{41 + 6 ln (2)}{72} \approx 0.6272$ $(41+6ln(2))/72~~0.6272$

Explanation:

We will operate on the assumptions $L > 0$ $L>0$ and that the probability is equal that $(a, b, c) = (x_{1}, x_{2}, x_{3})$ $(a, b, c) = (x_1, x_2, x_3)$ for all $(x_{1}, x_{2}, x_{3}) \in {[- L, L]}^{3}$ $(x_1, x_2, x_3) in [-L, L]^3$

Initial observations:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$ has real roots if and only if the discriminant of the quadratic is nonnegative, that is, $b^{2} - 4 a c \geq 0$ $b^2-4ac >= 0$ , or $b^{2} \geq 4 a c$ $b^2>=4ac$ .
We can partition the probability space into the two cases in which $sgn (a) = sgn (c)$ $"sgn"(a) = "sgn"(c)$ or $sgn (a) \neq sgn (c)$ $"sgn"(a)!="sgn"(c)$ , and
$P (sgn (a) = sgn (c)) = P (sgn (a) \neq sgn (c)) = \frac{1}{2}$ $P("sgn"(a)="sgn"(c)) = P("sgn"(a)!="sgn"(c)) = 1/2$
If $sgn (a) \neq sgn (c)$ $"sgn"(a) != "sgn"(c)$ , then $4 a c \leq 0$ $4ac <= 0$ , meaning $b^{2} \geq 4 a c$ $b^2>=4ac$ .
By symmetry:
$P (b^{2} \geq 4 a c ∣ a, c \leq 0) = P (b^{2} \geq 4 a c ∣ a, c \geq 0)$ $P(b^2>=4ac|a, c<=0) = P(b^2>=4ac|a, c>=0)$
As $b^{2} = {(- b)}^{2}$ $b^2=(-b)^2$ for all $b$ $b$ , we can restrict $b$ $b$ to $[0, L]$ $[0, L]$ without changing the probability of it falling within a certain range. Thus
$P (b^{2} \geq 4 a c ∣ a, c \geq 0) = P (b^{2} \geq 4 a c ∣ a, b, c \geq 0)$ $P(b^2>=4ac|a, c>=0) = P(b^2>=4ac|a, b, c>=0)$

With these, we can reformulate the problem as follows:

$P (a x^{2} + b x + c has real roots) = P (b^{2} \geq 4 a c)$ $P(ax^2+bx+c" has real roots") = P(b^2>=4ac)$

$= \frac{1}{2} P (b^{2} \geq 4 a c ∣ sgn (a) \neq sgn (c))$ $=1/2P(b^2>=4ac|"sgn"(a)!="sgn"(c))$

$+ \frac{1}{2} P (b^{2} \geq 4 a c ∣ sgn (a) = sgn (c))$ $+1/2P(b^2>=4ac|"sgn"(a)="sgn"(c))$

$= \frac{1}{2} (1) + \frac{1}{2} P (b^{2} \geq 4 a c ∣ a, c < 0 or a, c > 0)$ $=1/2(1)+1/2P(b^2>=4ac|a, c<0 or a, c >0)$

$= \frac{1}{2} + \frac{1}{2} P (b^{2} \geq 4 a c ∣ a, c > 0)$ $=1/2+1/2P(b^2>=4ac|a, c>0)$

$= \frac{1}{2} + \frac{1}{2} P (b^{2} \geq 4 a c ∣ a, b, c > 0)$ $=1/2+1/2P(b^2>=4ac|a, b, c>0)$

$= \frac{1}{2} + \frac{1}{2} P (b^{2} \geq 4 a c ∣ (a, b, c) \in {[0, L]}^{3})$ $=1/2+1/2P(b^2>=4ac|(a, b, c) in [0, L]^3)$

We will now calculate $P (b^{2} \geq 4 a c ∣ (a, b, c) \in {[0, L]}^{3})$ $P(b^2>=4ac|(a, b, c) in [0, L]^3)$

If we consider a coordinate system in 3-space, then ${[0, L]}^{3}$ $[0, L]^3$ is equivalent to a cube with side length $L$ $L$ resting in the first octant and having sides colinear with the axes. Let $S$ $S$ be the solid bounded by this cube and above by the surface $y^{2} = 4 x z$ $y^2 = 4xz$ , and $V_{S}$ $V_S$ be the volume of this solid. Then the probability that $b^{2} \geq 4 a c$ $b^2>=4ac$ given an arbitrary $(a, b, c) \in {[0, L]}^{3}$ $(a, b, c) in [0, L]^3$ is equal to the probability that $(a, b, c)$ $(a, b, c)$ falls within $S$ $S$ given that $(a, b, c)$ $(a, b, c)$ falls within the cube, i.e. $\frac{V_{S}}{V_{cube}} = \frac{V_{S}}{L^{3}}$ $V_S/V_"cube" = V_S/L^3$

To find $V_{S}$ $V_S$ , we will first consider the area $A_{b}$ $A_b$ of the slice of $S$ $S$ found by fixing $y = b$ $y=b$ for some $b \in [0, L]$ $b in [0, L]$ , and then we will add up all of these areas by integrating $A_{b}$ $A_b$ as $b$ $b$ goes from $0$ $0$ to $L$ $L$ .

If we fix $y = b$ $y=b$ for some $b$ $b$ , then $A_{b}$ $A_b$ is the area bounded by the lines $x = 0, x = L, z = 0, z = L$ $x=0, x=L, z=0, z=L$ and the curve $4 x z = b^{2}$ $4xz=b^2$ , which we can rewrite as $z = \frac{b^{2}}{4 x}$ $z=b^2/(4x)$

![desmos.com]( useruploads.socratic.org )

Notice that the curve intersects the square at the points where $x = L$ $x=L$ or $z = L$ $z=L$ , that is, at $(x, z) \in {(L, \frac{b^{2}}{4 L}), (\frac{b^{2}}{4 L}, L)}$ $(x, z) in{(L, b^2/(4L)), (b^2/(4L), L)}$

With that, we can now set up our integrals. The upper bound from $x = 0$ $x=0$ to $x = \frac{b^{2}}{4 L}$ $x=b^2/(4L)$ is the line $z = L$ $z=L$ . The upper bound from $x = \frac{b^{2}}{4 L}$ $x=b^2/(4L)$ to $x = L$ $x=L$ is the curve $z = \frac{b^{2}}{4 x}$ $z=b^2/(4x)$ . Thus

$A_{b} = \int_{0}^{\frac{b^{2}}{4 L}} L d x + \int_{\frac{b^{2}}{4 L}}^{L} \frac{b^{2}}{4 x} d x$ $A_b = int_0^(b^2/(4L))Ldx + int_(b^2/(4L))^Lb^2/(4x)dx$

and

$V_{S} = \int_{0}^{L} A_{b} d b$ $V_S = int_0^LA_bdb$

Omitting the full process of integration to save space, we find the result

$V_{S} = \frac{5 + 6 ln (2)}{36} L^{3}$ $V_S = (5+6ln(2))/36L^3$

Thus

$P (b^{2} \geq 4 a c ∣ a, b, c \in [0, L]) = \frac{V_{S}}{L^{3}} = \frac{5 + 6 ln (2)}{36}$ $P(b^2>=4ac|a, b, c in [0, L]) = V_S/L^3= (5+6ln(2))/36$

Substituting this into our initial equation, we get our final result:

$P (a x^{2} + b x + c has real roots)$ $P(ax^2+bx+c" has real roots")$

$= \frac{1}{2} + \frac{1}{2} P (b^{2} \geq 4 a c ∣ (a, b, c) \in {[0, L]}^{3})$ $=1/2+1/2P(b^2>=4ac|(a, b, c) in [0, L]^3)$

$= \frac{1}{2} + \frac{1}{2} (\frac{5 + 6 ln (2)}{36})$ $=1/2+1/2((5+6ln(2))/36)$

$= \frac{41 + 6 ln (2)}{72}$ $=(41+6ln(2))/72$

$\approx 0.6272$ $~~0.6272$

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Given ${a, b, c} \in [- L, L]$ ${a,b,c} in [-L,L]$ What is the probability that the roots of $a x^{2} + b x + c = 0$ $a x^2+b x + c = 0$ be real?

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Given ${a, b, c} \in [- L, L]$ ${a,b,c} in [-L,L]$ What is the probability that the roots of $a x^{2} + b x + c = 0$ $a x^2+b x + c = 0$ be real?