TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1990 question 1 solution
Prove that both 
and 
are non-negative for all real x.
Factorise to 
, which clearly is positive.
We can verify that is positive for all real x by rewriting it as 
, which clearly is positive.
The equation 
has clearly no real solutions for 
as we have earlier shown that is always positive, and -8(-x)+17 is always positive since two negative numbers multiplied is positive...
For the interval x>2 we can consider 
as we have earlier shown that always is positive for all real x. Factorise it to get 
. For positive x 
is always positive, and for numbers over 2
is always postitive (and when x=2 the whole equation is still positive, as we have the x^2 etc. terms as well.)
Now consider the interval 
. Factorise to 
, which clearly is positive. For 17-8x where x is less than 2 it is obvious that it will never be negative.
Prove that the equation 
has no real roots. First, consider the interval x<1, where we can use the factorisation into 
.
Now all squared things are clearly positive, as is the constant, and for the chosen interval 4(1-x) is positive as well.
Consider x=1, here everything stays the same, except the 4(1-x) part that disappears, and the expression as a whole is still positive.
Now consider x>1 - factorise into
(which is positive as the cube of a positive number is positive, and x-1 for x>1 is positive) and 
, which obviously is positive as it is an even power.
Therefore , for all x the equation has no real solutions (as all y-values are above the x-axis).
Solution by nota bene.
