How can I solve this nice rational equation

Question

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.

Answer 1

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 \begin{equation} \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} \end{equation} and \begin{equation} \label{eq5_30_06_2016} \dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}} \geqslant \dfrac{5}{2}-\dfrac{x}{2}. \end{equation} The first inequality equavalent to \begin{equation} \label{eq1_30_06_2016} 2(3 x^2+4 x+5) \geqslant (x+5)\sqrt{5 x^2+4 x+3}. \end{equation} 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 \begin{equation} \label{eq_2_30_06_2016} (x+1)^2\cdot \left(31 x^2-20 x+25\right) \geqslant 0. \end{equation} 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 $x=-1.$

Answer 2

Hint : Square the equation , isolate the term with the square-root and square the equation again.