Probable Real Quadratika.

The values of a and b a \ \text{and} \ b are equally possible in the square a 1 , b 1 |a| \leq 1, |b| \leq 1 . Find the probability (up to 3 decimal points) that the roots of the quadratic trinomial ( x 2 + 2 a x + b ) (x^2+2ax+b) are real.


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The answer is 0.667.

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1 solution

Shubham Garg
Jun 29, 2015

Let the values of a a be represented on x x and b b on y y axes respectively.

the area of square formed by condition given is 4 ( 2 × 2 ) (2×2) . ( x 1 , y 1 ) (|x| \leq 1 ,|y| \leq 1)

For the roots to be real a 2 b 0 a^2-b \geq 0 x 2 y \Rightarrow x^2 \geq y

the area bounded by parabola and square (outside the parabola) is 8 3 \frac 8 3 .

This is the favourable area.

P r o b a b i l i t y = f a v o u r a b l e a r e a t o t a l a r e a = 8 3 4 = 2 3 Probability = \frac{favourable \quad area}{total \quad area}=\frac {\frac 8 3 }{4} =\boxed { \frac 2 3}

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