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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1990 question 1 solution

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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.