ornate isle
So far I set z^9 = x, and rewrote it as 38x^2+bx+70, and then the factors would be (qx+r)(sx+t), so I also set up a system where qs = 38, qt+rs = b, and rt = 70. Then if you rearrange the first and third in the system as s = 38/q and t = 70/r you can plug in and get the middle one to be: b = 70(q/r) + 38(r/q). This is where I'm stuck. What do I do from here, if I've gone down the right path, or am I completely lost and is there a different, correct way to do it?
surreal kiteBOT
slim heath
So far I set z^9 = x, and rewrote it as 38x^2+bx+70, and then the factors would ...
this is a correct path
this is the eqation you have obtained
b = 70(q/r) + 38(r/q)
since (qx+r) is a factor, it satisfies the equation 38x^2+bx+70
hence x = -r/q will be a root. input that in the equation 38x^2+bx+70
you will obtain an eqn like 38t^2 - bt +70 = 0, where t = -r/q
obtain value of t by quadratic formula
put that in b = 70(q/r) + 38(r/q)
let me know if that worked
ornate isle
you will obtain an eqn like 38t^2 - bt +70 = 0, where t = -r/q
Wait, why does t = -r/q?
And also thanks for helping out on this
slim heath
oh well you can just go ahead with x = -r/q
my OCD ass is dead intent on making things look neat and oversimplified
ornate isle
Ohh instead of x you put t
slim heath
Ohh instead of x you put t
yeah
ornate isle
I was a little confused because t was in one of my roots
Okay thanks let me do that real quick
surreal kiteBOT
Okay thanks let me do that real quick
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ornate isle
Would b = 5320?
Wait hold on
Nevermind that's just the b^2 of the quadratic formula part lemme do the whole thing
So then just sqrt(5320)?
slim heath
can you show me the work?
slim heath
So then just sqrt(5320)?
i don't have anything to write atm