Find all possible solutions:
P=(x-1)*(x-2)*(x-4)*(x-8) =ax^2
Solve to get (x,y) for P(a)
Evaluate:
i. (x,y) for a=7 (original version)
ii. for a=4 (more friendly results!)
Credit: Valery Volkov
(In reply to
re(2): answer by Ady TZIDON)
I have also read your solution and found it normal and direct,
As to non-common approach , paying attention that 2*4=1*8
one might continue:
(x^2-9x+8)*(x^2-6x+8) =ax^2
What? Literally the second and third lines in my solution are:
Group the left side and multiply (x-2)*(x-4) together and (x-1)*(x-8) together:
(x^2-9x+8) * (x^2-6x+8) = ax^2
and then continue with z=x^2+8
After that I do not see the value in substituting z=x^2+8, since then you'd get something like (z-9x)*(z-6x) = a(z-8). How is this helping to make things simpler?
Rather, go further down to
x^2-(15/2)x+8 = +/- sqrt[a+9/4]*x
The presence of sqrt[a+9/4] is what makes an answer "ugly". While in the queue I suggested a=10 and Jer suggested a=4. Both these choices reduce sqrt[a+9/4] to a rational number. Then the roots of the polynomial are algebraic of degree 2; which is simpler than most arbitrary values of a, including a=7, where the roots are algebraic of degree 4.
So if you wanted to make a substitution for clarity, then take z=sqrt[a+9/4] or z=sqrt[4a+9].
Going with z=sqrt[4a+9]. Then I will have solutions go from
x = (15-sqrt[4a+9]+sqrt[4a+106-30sqrt[4a+9]])/4
to
x = (15-z+sqrt[z^2-30z+97])/4
re(3): answer
