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3 roots of a biquadratic (Posted on 2021-03-14) Difficulty: 3 of 5
Real numbers a and b are chosen so that each of two quadratic trinomials x2+ax+b and x2+bx+a has two distinct real roots and the product of these trinomials has exactly three distinct real roots.

Determine all possible values of the sum of these three roots.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution | Comment 1 of 3
a cannot equal b, else there would be only two distinct real roots.

Solve each quadratic using the quadratic formula to yield:
x = (-a+/-sqrt(a^2-4b))/2
x = (-b+/-sqrt(b^2-4a))/2

To have three distinct roots overall, one of the smaller roots of one quadratic must equal the larger root of the other quadratic.  Wlog:
(-a-sqrt(a^2-4b))/2 = (-b-sqrt(b^2-4a))/2

Rearrange a bit to get the radicals on one side:
b-a = sqrt(a^2-4b) + sqrt(b^2-4a)

Square both sides and rearrange:
2a + 2b - a*b =  sqrt(a^2-4b) * sqrt(b^2-4a)

Square both sides and rearrange again:
a^3 - a^2*b - a*b^2 + b^3 + a^2 - 2a*b + b^2 = 0

This last equation factors into (a+b+1) * (a-b)^2 = 0
The solution of a=b must be discarded leaving a+b+1=0 as a necessary condition for a and b.

Apply this to the two quadratics to get x^2 + ax - (a+1) and x^2 + bx - (b+1).  These quadratics factor into (x-1) * (x+a+1) and (x-1) * (x+b+1).

Then the common root is x=1.  Then the sum of the roots is (-a-1) + 1 + (-b-1) = -1*(a+b+1) = -1*0 = 0.  The sum of the three distinct roots is always 0 for any valid choice of a,b.

  Posted by Brian Smith on 2021-03-14 17:03:52
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