Solve the inequality $ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $

Question

Solve the inequality $$ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $$

---

Attempt:

I get that $2x^{2} + x + 4 > 0$ for all $x$. Let $y = 2x^{2} + x$, then the inequality becomes $$ y - 4 \sqrt{y + 4} < 1 $$ $$ 0 < \frac{1 + 4 \sqrt{y+4} - y}{y}$$ so from here I get that $0 < y < 9 - 2\sqrt{14} \: \:\cup \:\: y > 9 + 2\sqrt{14}$. This is for $y>0$ or $x(2x+1)>0 \implies x > 0 \:\: \cup \:\: x < -1/2$

I may continue to solve for $x$.. but the answer key says that the solution is $ -7/2 < x < 3$. I doubt that the solution set will be that tidy. Are there better approaches to solve the inequality?

Answer 1

Let $y= \sqrt{2x^{2} + x + 4} \geq 0$, (notice that $2x^2+x+4>0$ for all real $x$) then we get $$ y^2-4-4y<1\implies y^2-4y-5 = (y-5)(y+1)<0$$

So $$y\in (-1,5)$$ and so $$ \sqrt{2x^{2} + x + 4} < 5 \implies 2x^2+x+4<25$$

so $$2x^2+x-21= (2x+7)(x-3)<0 \implies x\in (-{7\over 2},3)$$

Answer 2

It's $$2x^2+x+4-4\sqrt{2x^2+x+4}+4<9$$ or $$\left(\sqrt{2x^2+x+4}-2\right)^2<9$$ or since $\sqrt{2x^2+x+4}>0$, it's $$\sqrt{2x^2+x+4}<5$$ or $$2x^2+x-21<0$$ or $$(x-3)(2x+7)<0$$ or $$-3.5<x<3.$$