There doesn't seem to be much actual "work". ?
If a,b, and c are integers, you are supposed to know that if [imath]g+\sqrt{h}[/imath] is a solution, then so is [imath]g-\sqrt{h}[/imath] . You do have three equations.Your example => a, b
My problem=> a,b,c
But there are only 2 equations.
I think this concept is used only when g+root hIf a,b, and c are integers, you are supposed to know that if [imath]g+\sqrt{h}[/imath] is a solution, then so is [imath]g-\sqrt{h}[/imath] . You do have three equations.
-Dan
Why do you want to query what you've been told and come up with some outlandish extrapolation?I think this concept is used only when g+root h
g is rational number and h is irrational number
why p(7) = p(5+2) can we get p(5-2)
or p(7) = p(4+3) can we get p(4-3)
Yes, I should have been more specific here.I think this concept is used only when g+root h
g is rational number and h is irrational number
why p(7) = p(5+2) can we get p(5-2)
or p(7) = p(4+3) can we get p(4-3)
Where are you getting that [imath]p(-7)[/imath] is also equal to zero? Where are you getting that [imath]p\left(-1+\sqrt{3\;}\right)[/imath] is also equal to zero?
Because P(x) has only even degreeWhere are you getting that [imath]p(-7)[/imath] is also equal to zero? Where are you getting that [imath]p\left(-1+\sqrt{3\;}\right)[/imath] is also equal to zero?
Where are you getting that [imath]-1 \pm \sqrt{3\;}[/imath] are factors?
Are you familiar at all with the Quadratic Formula?
Because P(x) has only even degreesBecause P(x) has only even degree
x^6, x^4, x^2 so we can get p(-x) = p(x).