Imagination not allowed

Algebra Level pending

Parameter $a$ is chosen randomly and uniformly in the interval $(-1,2)$ .

What is the probability that the given equation will have all of it's roots real?

$x^4 +2ax^3 +(2a-2)x^2+(-4a+3)x-2=0$

The answer is in the form of $\frac {a}{b}$ , where $a,b$ are co-prime positive integers. Find $|a^2-b^2|$ .

The answer is 8.

1 solution

Parth Sankhe
Oct 21, 2018

Separating the terms containing $a$ and those who do not, we get

$(x^4-2x^2+3x-2)=-2a(x^3+x^2-2x)$

The RHS of the equation gets factorised into $x(x-1)(x-2)$ . Coincidentally, x=1, 2 are also solutions of the LHS. By factorising and eliminating we get:

$(x^2-x+1)+2ax=0$

We already have 2 real roots (1,2), now we need the remaining equation to have real roots as well. Thus, it's discriminant has to be non-negative.

Putting discriminant $>0$ we get another quadratic equation in $a$ .

$4a^2-4a-3>0$

$a>\frac {3}{2}$ and $a<-½$

Thus the satisfied length of interval would be

$\frac {(-½-(-1))+(2-\frac {3}{2})}{2-(-1))}=\frac {1}{3}$

Hence, the answer is $|3^2-1|=8$

Imagination not allowed

The answer is 8.

1 solution

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