Answer the question below.?

Question

If you have x^2+a*x+b=0, solve it to get two roots a1<=b1.
Now, form another similar equation: x^2+a1*x+b1=0.
Solve this second equation to get two roots a2<=b2.
Continue this pattern.

Question: Prove that you can not produce more than five equations. That is, prove that you will reach, in the best case, up to x^2+a5*x+b5=0, which will not have solutions in real number system.

Note: a and b are nonzero real numbers.

Answer 1

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Answer 2

You did not constrain the problem sufficiently. There is a trivial solution that proves the theorem false. let a=b=0 and there are an infinite number of iterations possible.
a1=b1=0
a2=b2=0
..........
an=bn=0
All the formulas become X^2=0

Answer 3

I guess you'll just have to brute-force it. Use the quadratic formula for each case and you'll get an equation which cannot be true for any real number the fifth time around.
Here's the first 2:
x^2 + ax + b =0
a1 = (-a-root(a^2-4a))/2, b1 = -a+root...
x^2 + a1x + b1 = 0
a2 = (root(2((a-4)root(a^2-4b)+ a^2 + 4a-2b))-root(a^2-4b)-a)/4
b2 = (root(2((a-4)root(a^2-4b)+ a^2 + 4a-2b))+root(a^2-4b)-a)/4
Good luck finding the next 3...;)

Answer 4

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