Solve this inequality: $\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5}$.
My solution:
Multiply $(x+5)^2$ on both sides:
$\frac{(x^2 -2)(x+5)^2}{2} = \frac{x^4 + 10x^3 + 23x^2 -20x -50}{2}$
$(6x^2 -8x - 1)(x+5) = 6x^3 + 22x^2 -41x - 5 $
$\frac{x^4 + 10x^3 + 23x^2 -20x -50}{2} < 6x^3 + 22x^2 -41x - 5$
$0.5x^4 - x^3 - 10.5x^2 + 31x - 20 < 0 $
I don't think doing this way will lead to solving the question....
$\endgroup$ 23 Answers
$\begingroup$Solution ($\color{red}{Edited}$):
Given
$$\frac{(x^2 -2)}{2} < \frac{(6x^2 -8x - 1)}{(x+5)}$$1. Multiply both sides by 2$(x+5)^2$ ($\color{red}{Since\;}$$\color{red}{(x+5)^2\ge 0,\;the\;inequality\;is\;preserved}$):$$(x+5)^2(x^2-2)<2(x+5)(6x^2-8x-1)$$2. Expand and collect like terms:$$x^4-2x^3-21x^2+62x-40<0$$3. Factor the polynomial:$$(x+5)(x-1)(x-2)(x-4)<0$$4. Draw the critical points on the real line
We now can write the domain as the union $$D=D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5=(-\infty,-5) \cup (-5,1) \cup (1,2) \cup (2,4) \cup (4,\infty)$$6. By studying the sign of each factor, answer the following: Is the left-hand side negative (less than zero)?
- On $D_1$: $$ (-)(-)(-)(-)=(+)<0$$ which is false.
- On $D_2$: $$ (-)(-)(-)(+)=(-)<0$$ which is true.
- On $D_3$: $$ (+)(-)(-)(+)=(+)<0$$ which is false.
- On $D_4$: $$ (+)(+)(-)(+)=(-)<0$$ which is true.
- On $D_5$: $$ (+)(+)(+)(+)=(+)<0$$ which is false.
Thus the solution is $$x=(-5,1)\cup (2,4)$$
as shown in the following plot
HINT: use that $$\frac{6x^2-8x-1}{x+5}-\frac{x^2-2}{2}=\frac{(4-x)(x-2)(x-1)}{2(5+x)}$$
$\endgroup$ 2 $\begingroup$This is very similar to Meshal's approach, but you can avoid having to factor a quartic. It gives the general idea for a slightly different approach as well.
Move all terms to the same side: $$\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5} \iff \frac{x^2 -2}{2} - \frac{6x^2 -8x - 1}{x+5} < 0$$ Now put everything on a common denominator and factor the cubic in the numerator: $$\frac{x^3-7 x^2+14 x-8}{2(x+5)} < 0 \overset{\;\color{green}{(*)}}{\iff} \frac{(x-1) (x-2) (x-4)}{2(x+5)} < 0$$ You can make a sign table of this fraction which leads immediately to the solution set and/or follow a scheme like Meshal's.
You can also notice that all roots (three in the numerator and one in the denominator) are simple roots (i.e. their multiplicity is 1) and thus sign changes occur in each root.
These roots are in increasing order $-5$, $1$, $2$ and $4$ and clearly the inequality is not satisfied for $x<-5$ (check with a value or reason for $x \to -\infty$); so it is satisfied for $x \in (-5,1) \cup (2,4)$.
To factor in step $\color{green}{(*)}$, you can follow general algorithms or group as follows: $$\begin{align} \color{blue}{x^3}\color{red}{-7 x^2+14 x}\color{blue}{-8} & = \color{blue}{(x-2)(x^2+2x+4)}\color{red}{-7x(x-2)} \\ & = (x-2)(x^2-5x+4) \\ & = (x-2)(x-1)(x-4) \end{align}$$
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