You are not logged in.
Pages: 1
a) Use algebra to find the set of values of x for which x(x-5) > 36
b) Using your answer to part a, find the set of values of y for which y²(y²-5) > 36
Thanks.
Last edited by Daniel123 (2007-10-22 00:49:06)
Offline
x(x-5)>36
x² - 5x - 36 > 0
(x-9)(x+4) > 0.
Drawing a sketch of the graph shows that the solution to the inequality is then x<-4 or x>9.
For the second part, we now need to find values of y such that y²<-4 or y² >9.
Squares are never negative, so nothing will fit the first part. y² > 9 can be solved though, and has the solution y<-3 or y>3.
(Or, if you want to be clever, |y|>3.)
Why did the vector cross the road?
It wanted to be normal.
Offline
aah, it was the y²<-4 that threw me. Thanks.
Offline
Pages: 1