Let P(x) and Q(x) be (monic) polynomials with real coefficients (the first coefficient being equal to 1), and degP(x)=degQ(x)=10.
Prove that if the equation P(x)=Q(x) has no real solutions, then P(x+1)=Q(x-1) has a real solution.
If P(x)=Q(x) has no solutions then P(x)-Q(X) cannot be an odd polynomial. This means their x^9 terms have the same coefficient. If they didn't, this difference would be an odd degree polynomial which will have at least one real root.
P(x)=x^10 + ax^9 + other smaller degree terms.
Q(x)=x^10 + ax^9 + other smaller degree terms.
P(x+1)=(x+1)^10 + a(x+1)^9 + other smaller degree terms.
=x^10 + (10+a)x^9 + other smaller degree terms.
Q(x-1)=(x-1)^10 + b(x-1)^9 + other smaller degree terms.
=x^10 + (-10+a)x^9 + other smaller degree terms.
If we equate these and find the difference we get
P(x+1)-Q(x-1)=20x^9 + other smaller degree terms.
Which, being an odd polynomial, will have at least one root.
|
|
Posted by Jer
on 2021-03-20 12:01:27 |
Solution