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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.

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  • $\begingroup$ $$\sqrt{5x^2+4x+3}=a,\sqrt{10x^2+9x+8}=b\implies b^2-2a^2=x+2$$ and $$3x^2+4x+5=\dfrac{3a^2+8(b^2-2a^2)}5, 8x^2+9x+10=\dfrac{4b^2+9(b^2-2a^2)}5$$ $\endgroup$ Jun 29 '16 at 9:06
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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 $x=-1.$

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

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  • $\begingroup$ 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. $\endgroup$ Jun 29 '16 at 9:17
  • $\begingroup$ True, but I do not see another possibility to solve the equation. But anyway, it is easy to check which solutions satisfy the original equation. It is well known that squaring is not an equivalence transformation, but at least we do not lose solutions by squaring. $\endgroup$
    – Peter
    Jun 29 '16 at 9:24
  • $\begingroup$ I fully agree with you, be sure ! I tried a few things .... without any success ! Cheers. $\endgroup$ Jun 29 '16 at 9:29

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