Submitted by maths123 on Mon, 05/25/2015 - 18:09
Forums:
For the last part of the question, I replicated the earlier parts and got to the below stage when calculating the root locus:
$ y=0 \implies p = \frac{-x^2 \pm \sqrt (x^4 - 8x) } {2x} $
From here I proceeded as follows,
since p is a number and is real we have two deductions:
(1) the denominator is not equal to 0 and hence x is not equal to 0 (or else p is undefined)
(2) p is real so the discriminant is greater than or equal to 0
$ x^4 - 8x \geq 0 $
from deduction (1) we know the x is not equal to 0 so I divided by x
$x^3 - 8 \geq 0 $
$x \geq 2 $
the answers say that $ x<0$ or $x \geq 2 $ - I agree with the answers but what is wrong with the deduction process I have used above.
The problem lies when when
The problem lies when when you divided by $x$. Unlike equalities, you cannot simply divide by algebraic factors in an inequality unless you know their sign (in this case $x$ could be positive or negative and if you divide an inequality by a negative quantity you have to reverse the inequality). If you are trying to solve $$x^4-8x\geq 0,$$ it is often easiest to make a sketch of the graph $y=x^4-8x$ and see where $y$ is greater than 0.