(i)
We have:
\displaystyle \frac {x + 2x^2 - x^3} x - \frac 2 x > 0
That is:
\displaystyle \frac {-(x^3 - 2x^2 - x + 2)} x > 0
We can rewrite this as:
\displaystyle x(x^3 - 2x^2 - x + 2) < 0
by multiplying by -x^2.
It remains to factorise this cubic. By inspection, we see that x = 1 is a root, since 1 - 2 - 1 + 2 = 0, so we expect:
x^3 - 2x^2 - x + 2 = (x - 1)(x^2 + Ax + B)
for constants A,B. We see that B = -2 and A - 1 = -2, so A = -3. So:
x^3 - 2x^2 - x + 2 = (x - 1)(x^2 - 3x - 2) = (x - 1)(x + 1)(x - 2)
So we’re looking at:
(x + 1)x(x - 1)(x - 2) < 0
It helps to look at a sketch of y = (x + 1)x(x - 1)(x - 2):
(we know that it’ll look like this since this a quartic with positive leading coefficient)
We can then see that either -1 < x < 0 or 1 < x < 2.
(ii)
Note that this inequality is equivalent to:
\left|\sqrt {3x + 10}\right| > \left|2 + \sqrt {x + 4}\right|
Squaring:
3x + 10 > (2 + \sqrt {x + 4})^2
expanding:
3x + 10 > 4 + 4 \sqrt {x + 4} + x + 4
so that:
0 > -2x + 4 \sqrt {x + 4} - 2
or, easier to deal with:
2x - 4 \sqrt {x + 4} + 2 > 0
This is a “hidden quadratic” in \sqrt {x + 4}. We’ll have:
A(x + 4) + B\sqrt{x + 4} + C = 2x - 4 \sqrt {x + 4} + 2
We can see that A = 2 and B = -4 off the bat, then we need 4A + C = 2, so 8 + C = 2, giving C = -6. So the inequality is equivalent to:
2(x + 4) - 4 \sqrt {x + 4} - 6 > 0
dividing by 2:
(x + 4) - 2 \sqrt {x + 4} - 3 > 0
Note that we can factorise this:
(\sqrt {x + 4} - 3)(\sqrt {x + 4} + 1) > 0
(to make this easier to see, note that we can write u = \sqrt {x + 4} then the LHS becomes u^2 - 2u - 3)
Since \sqrt {x + 4} + 1 > 0 for all x, the inequality holds if and only if \sqrt {x + 4} - 3 > 0. That is, \sqrt {x + 4} > 3, equivalent to x + 4 > 9. So the inequality holds for \boxed{x > 5}
