# How can I solve this nice rational equation

I am trying solve this equation $$\dfrac{3x^2 + 4x + 5}{\sqrt{5x^2 + 4x +3}}+\dfrac{8x^2 + 9x + 10}{\sqrt{10x^2 + 9x +8}} = 5.$$ Where $x \in \mathbb{R}$. I knew that $x=-1$ is a given solution. But I can't solve it. I tried We rewrite the given equation in the form $$\dfrac{3(x^2 + x + 1) + x + 2}{\sqrt{5(x^2 + x + 1) - (x+2)}}+\dfrac{8(x^2 + x + 1) + x + 2}{\sqrt{10(x^2 + x + 1)- (x+2)} } = 5.$$ Put $a = x^2 + x + 1$ and $b = x + 2$, we get $$\dfrac{3a + b}{\sqrt{5a - b}}+\dfrac{8a + b}{\sqrt{10a- b} } = 5.$$ From here, I stoped.

• $$\sqrt{5x^2+4x+3}=a,\sqrt{10x^2+9x+8}=b\implies b^2-2a^2=x+2$$ and $$3x^2+4x+5=\dfrac{3a^2+8(b^2-2a^2)}5, 8x^2+9x+10=\dfrac{4b^2+9(b^2-2a^2)}5$$ Jun 29 '16 at 9:06

The equation of tangent line of the grap of the funtion $y = \dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}$ at the point $x=-1$ is $y=\dfrac{x}{2}+\dfrac{5}{2}$ and the equation of tangent line of the grap of the funtion $y = \dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}$ at the point $x=-1$ is $y=\dfrac{5}{2}-\dfrac{x}{2}$. We prove that $$\label{eq4_30_06_2016} \dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}} \geqslant \dfrac{x}{2}+\dfrac{5}{2}$$ and $$\label{eq5_30_06_2016} \dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}} \geqslant \dfrac{5}{2}-\dfrac{x}{2}.$$ The first inequality equavalent to $$\label{eq1_30_06_2016} 2(3 x^2+4 x+5) \geqslant (x+5)\sqrt{5 x^2+4 x+3}.$$ If $x+5 \leqslant 0$, it is always true.
If $x+5 > 0$, squaring both sides , we get $$31 x^4+42 x^3+16 x^2+30 x+25 \geqslant 0.$$ Equavalent to $$\label{eq_2_30_06_2016} (x+1)^2\cdot \left(31 x^2-20 x+25\right) \geqslant 0.$$ Similarly, if $x \geqslant 5$ the second equality is true. If $x < 5$, squaring both sides, we get $$(x+1)^2 \cdot\left(246 x^2+175 x+200\right) \geqslant 0.$$ Equality of two equalities hold iff $x = -1$.
Add two equalities, we have $$\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}+\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}\geqslant 5.$$ Therefore, the equation $$\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}+\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}= 5$$ has only solution $x=-1.$
• The problem is that we should end with a polynomial of degree $10$ (after removing the $(x+1)^2$ common factor) and that successive squaring introduces extra roots while the original problem has $5$ solutions. Jun 29 '16 at 9:17