Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$

Question

Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$

---

This is my try

Let $t=\sqrt{2x+3}-2\sqrt{x+1}$ or $t^2=6x+7-4\sqrt{2x^2+5x+3}=1$

The equation is equivalent to: $t^2-4(x+2)t-6x-3=0\qquad(*)$

I have no idea how to solve the equation $(*)$. Who can help me?

Answer 1

$$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$$ $$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})=1-\sqrt{2x^2+5x+3}$$ $$(x+2)^2(6x+7-4\sqrt{(2x+3)(x+1)})=2x^2+5x+4-2\sqrt{2x^2+5x+3}$$ $$(x+2)^2(6x+7)-(2x^2+5x+4)=2(2(x+2)^2-1)\sqrt{(2x+3)(x+1)})$$ $$(6 x^3+29 x^2+47 x+24)^2=4(2x^2+8x+7)^2(2x+3)(x+1)$$ Latter can be factored as roots $-1,\,-\frac{1}{2},3$ are known.
$\frac{4 x^6+12 x^5-19 x^4-106 x^3-135 x^2-68 x-12 }{(x+\frac{1}{2})(x+1)(x-3)}$$=2 (2 x+1) (x+2)^2$

Answer 2

$$ (x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1 $$ Let $t=x+1$, $$2x+3=2t+1$$ $$(t+1)(\sqrt{2t+1}-2\sqrt{t})+\sqrt{2t^2+t}=1$$ $$(t+1)(\sqrt{2t+1}-2\sqrt{t})=1-\sqrt{2t^2+t}$$ Squaring both sides, $$(t^2+2t+1)(2t+1+4t-4\sqrt{2t^2+t})=1+2t^2+t-2\sqrt{2t^2+t}$$ $$6t^3+12t^2+6t+t^2+2t+1-\sqrt{2t^2+t}(-4t^2-8t-2)=1+2t^2+t$$ $$t(6t^2+11t+7)=\sqrt{t}\sqrt{2t+1}(-4t^2-8t-2)$$ Again Squaring both sides, $$ 4t^6-12t^5-19t^4+10t^3+9t^2-4t=0 $$ Therefore , $t=-1,0,\frac{1}{2},4$ $$ x=-2,-1,-\frac{1}{2},3 $$