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STEP II 1990 question 1 solution

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

Prove that both x^4-2x^3+x^2 and x^2-8x+17 are non-negative for all real x.

Factorise x^4-2x^3+x^2 to (x^2-x)^2, which clearly is positive.

We can verify that x^2-8x+17 is positive for all real x by rewriting it as (x-4)^2+1, which clearly is positive.


The equation x^4-2x^3+x^2-8x+17=0 has clearly no real solutions for x\le0 as we have earlier shown that x^4-2x^3+x^2 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 x^4-2x^3 as we have earlier shown thatx^2-8x+17 always is positive for all real x. Factorise it to get x^3(x-2). For positive x x^3 is always positive, and for numbers over 2(x-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 0<x\le2. Factorise x^4-2x^3+x^2 to (x^2-1)^2, 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 x^4-x^3+x^2-4x+4=0 has no real roots. First, consider the interval x<1, where we can use the factorisation into 4(1-x)+x^2((x-\frac{1}{2})^2+\frac{3}{4}).

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 intox^3(x-1) (which is positive as the cube of a positive number is positive, and x-1 for x>1 is positive) and (x-2)^2, which obviously is positive as it is an even power.


Therefore , for all x the equation x^4-x^3+x^2-4x+4=0 has no real solutions (as all y-values are above the x-axis).

Solution by nota bene.

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