Solve the equation $10x^2+11x+4=8(x+1)\sqrt{2x^2+x-1}$

Question

Sovle the equation $10x^2+11x+4=8(x+1)\sqrt{2x^2+x-1}$.

I tried put $\sqrt{2x^2+x-1}= t,\; (t \geq0)$, so we have : ${2x^2+x-1}= t^2$ and $10x^2+11x+4=8(x+1)t$.

Answer 1

Given $$10x^2+11x+4=8(x+1)\sqrt{2x^2+x-1}$$ If we divide both sides by $8(x+1)$, square both sides, and subtract the right side, we get

$$\bigg(\frac{10 x^2 + 11 x + 4}{8 (x + 1)}\bigg)^2 - \bigg(\sqrt{2 x^2 + x - 1}\bigg)^2 = 0$$ Expanding, we get $$28 x^4 + 100 x^3 - 9 x^2 - 152 x - 80 = 0$$

There is a quartic formula but Wolfram Alpha gives and answer here as

$$\quad x≈-3.2368\quad x≈1.2880 \qquad x≈-0.81127 - 0.16488 i \qquad x≈-0.81127 + 0.16488 i $$