Need help to solving the inequality term of $\frac{|2x-4|}{5x-15} \geq \frac{2+\sqrt{x-8}}{x^2+3x-130}$

Question

$$\frac{|2x-4|}{5x-15} \geq \frac{2+\sqrt{x-8}}{x^2+3x-130}$$ My work so far is :

Finding the definition term for the left hand side. $$5x-15 \neq 0 $$ $$5(x-3)\neq 0 $$ $$x \neq 3$$

Finding the definition term for the right hand side. $$x^2+3x-130 \neq 0$$ $$(x+13)(x-10) \neq 0$$ $$x\neq -13 \lor x\neq 10$$

How to finding the inequality term and simplify this form ?

I also try raising both side into power of two, but i fail here $$ (2+\sqrt{x-8} )^2$$ This still contain radical form.

Please help me to solving this inequality. Thank you so much.

Answer 1

Clearly $x\geq 8$. Let $t=\sqrt{x-8}$, so $t\geq 0$. Then $x= t^2+8$ and we can rewrite the inequality as:

$${2(t^2+6)\over 3(t^2+5)}\geq {2+t\over (t^2+21)(t^2-2)}$$

If $t<\sqrt{2}$, i.e. $x<10$ then this is always true (since the left side is positive and right is negative).

So we have to consider only the case when $t>\sqrt{2}$. In this case we get a following polynomial inequality we have to solve:

$$2(t^2+6)(t^2+21)(t^2-2)\geq 3(t^2+5) (2+t)$$

Answer 2

For $x>10$ we have $$\frac{(x+13)(2x-4)(x-10)}{3x+15}-2=\frac{2 \left(x^3+x^2-139 x+245\right)}{3 (x+5)}>0$$ for $x>10$. So we can square both sides and we get: $$\frac{4 x^6+8 x^5-1108 x^4+839 x^3+79226 x^2-271945 x+241900}{9 (x+5)^2}\geq 0$$ This is true for $$10.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000<x\leq 10.06823272213296295839689196005349825199611352422267344083408144974074308711520673149338 347871794717\lor x\geq 10.41706305014369586123824673769643174365918019989949426356238792171774902413061605903142 407998151590$$ It remaines the case $$8\le x<10$$