Non-Linear Inequality.

p^2-7p+10<0

solve the roots for the inequality equaling zero, using the good old

(-b +/- sqrt(b^2 - 4ac))/2a
(7 +/- sqrt(49 - 4 * 10)) /2 = (7 +/- 3)/2 = 5 or 2 as the outliers
2*2 - 7*2 +10 = 0, 5*5 -35+10 =0

So aren't 2 and 5 your boundaries by this logic?


-(2-1)^0.5 = -1 vs 2-3 = -1 looks like a boundary and 2 < 3! also
-(5-1)^0.5 = -2 vs 5-3 =2 => doesn't work - as you alluded in your boundary conditions (hint as the signs are reversed you have a -ve mistake somewhere (when you squared negatives and didn't reverse the >)!,

You believed the left side is negative, therefore the right side is also negative, so upon squaring both sides, the inequality sign should not change because both sides are multiplied by a negative, right?

-1 > -3 , and both sides are negative, but square both sides and you get
1 > 9 not so, see your mistake!

So your correct lower boundary should have been 1 and your upper boundary is 2.

1 < p < 2.

That's easy to state - but how to prove?

Hint let x = p-1, substitute this in to get:

- sqrt x > x -2 (now times by -1 and reverse the > ) and check the boundaries
sqrt x < 2 - x (now lets apply some simple logic to get gross boundaries)

first 0 < sqrt x (implies x is positive is the lower boundary, and this in turn implies)

2 - x < 2 (as x is positive), so put this together and you get

0 < sqrt x < 2 - x < 2

Now the lower boundary is a clear hard boundary x > 0, lets see what this means for p, as p = x + 1 then x + 1 > 0 +1 so substituting back in for p gives p > 1 as the lower boundary.

Now check the higher boundary and can you see x < 2, but in fact the constraint is tighter than this ( x << 2) in that the sqrt x < 2 -x, the transition point is actually a lower boundary of x < 1 (not 2), so p has an upper limit of being p < 2.

Can you see why?

sqrt(x) < (2-x) so square both positive sides as you suggested
x < 4 - 4x + x^2
0 < 4 - 5x + x^2

Use the quadratic roots formulea to get (5 +/- (25 - 4*4)^0.5)/2 = (5 + / - (9)^0.5)/2 = (5 +- 3) /2

To give possible solutions x = 1 or 4, but apply the first limit of x < 2 and you can discard 4, leaving only 1 as the hard upper limit and if x < 1 then x + 1 < 1 + 1 so p must be < 2.

So finally 1 < p < 2

Negative roots in non-linear inequalities are very tricky!

Thank-you very much, once again!

So what is the principle when squaring an inequality where both sides are negative? I mean, the only rule pertaining to inequalities is that the sign only changes when multiplying by a negative, so by what principle does one reverse the sign when squaring and both sides are negative, besides doing such a thing:
-1>-3 thus
1<9 ?
Is there some sort of mathematical logic?

But thank you very much for your help!

Squaring both sides of an non-linear inequality when you have both sign and magnitude changes is complex.

1. If both sides are always positive the inequality doesn't change.
2. If both sides are always negative the inequality reverses.
3. If one side is positive and the other negative the inequality MAY flip depending on magnitude alone which dominates, and so matters amongst these rules becomes tricky where an inflection point comes along as the equation is non-linear.

Simple examples for a linear, constant.

If 2 < 4, then squaring 4 < 16
If -4 <2> 4

-4 < 8, squared 16 < 64, but
-8 <4> 16 (magnitude dominates)

Now in your example I had to sit for about 10 - 15 minutes to think that through as all your working was clear and well thought out, but just plugging the values 0 -> 5 into the equation showed something was wrong. This really forced a detailed think of what was going wrong, not just squaring an inequality with one point where you might get a transitional sign flip and another point where it would skew into complex numbers. This was a challenging problem that needed both simplifying, and really carefully checking of transition points, limits, sign changes and assumptions of the upper limit.

It would be very easy to get this one wrong - if you didn't field check your results as you did - a very good technique to apply. And even when you see the deeper parts of the problem as you said its exacting to work out how to get the right answer, even when with Excel you can see the correct answer in under 3 minutes!

Glad to help out.

I guess then it has something to do with absolute values, but I don't want to go there!

But thank you very much, your always a great help!