What is the number of REAL solutions of the following equation?

(9/10)^k=-3+x-x^2 where k belongs to the set of real numbers?

1 Answer
May 17, 2018

0

Explanation:

First off, the graph of a^x, a>0 will be continuous from -ooto+oo and will always be positive.

Now we need to know if -3+x-x^2>=0

f(x)=-3+x-x^2

f'(x)=1-2x=0

x=1/2

f''(x)=-2<- so the point at x=1/2 is a maximum.

f(1/2)=-3+1/2-(1/2)^2=-11/4

-3+x-x^2 is always negative while (9/10)^x is always positive, they will never cross and so have no real solutions.