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I came across this equation: $x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$

Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing the fractions together and then squaring. Is there any?

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As $x>0$

WLOG $x=3\sec t, 0< t<\dfrac\pi2\implies \sin t,\cos t>0$

$$\dfrac{35}4=3\sec t+\dfrac{3\sec t}{3\tan t} \iff\sec t+\csc t=\dfrac{35}{12}$$

Method$\#1:$

Let $\sin t+\cos t=u$

$$\dfrac{35}{12}=\dfrac{2u}{u^2-1}$$

$$\iff0=35u^2-24u-35=7u(5u-7)+5(5u-7)=(5u-7)(7u+5)$$

As $u>0, u=\sin t+\cos t=\dfrac75$

Now use $(\sin t+\cos t)^2+(\cos t-\sin t)^2=2$

$\cos t-\sin t=\pm\sqrt{2-\left(\dfrac75\right)^2}=\pm\dfrac15$

$2\cos t=\cos t-\sin t+\cos t+\sin t=\dfrac{7\pm1}5$

Method$\#2:$

As $\sec^2t+\csc^2t=\sec^2t\csc^2t$

Let $u=\sec t\csc t=\dfrac2{\sin2t}$

Squaring we get $$\left(\dfrac{35}{12}\right)^2=u^2+2u\iff(u+1)^2=\left(\dfrac{37}{12}\right)^2\implies u=\dfrac{25}{12}\text{ as }u>0$$

$\sin2t=\dfrac2u=\dfrac{24}{25}\implies\cos2t=\pm\dfrac7{25}$

$x=\dfrac3{\cos t}=+\dfrac6{\sqrt{2(1+\cos2t)}}=?$

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We see that we need $x>3$ then let $x=\frac3{\cos y}$ with $y\in\left(0,\frac \pi 2\right)$

$$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4} \iff \frac1{\cos y}+\frac1{\sin y}=\dfrac{35}{12}$$

and by half tangent identities by $t=\tan \frac y2$ we obtain

$$\frac{1+t^2}{1-t^2}+\frac{1+t^2}{2t}=\dfrac{35}{12} \iff (3t-1)(2t-1)(t^2-7t-6)=0$$

and $t=\frac 13, \frac12$ lead to the answer indeed

  • $\dfrac3{\cos(2\cdot \arctan\left(\frac13\right)}=5$
  • $\dfrac3{\cos(2\cdot \arctan\left(\frac12\right)}=\frac{15}4$
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It' not a coincidence. Let $f(x)=x + g(x)$ where $g(g(x))=x$, that is, $g$ is an involution. We'll show that $f(x)=f(a-x)$, provided that $f(x)=a$.

Assume $f(x)=a$, that is $a-x=g(x)$. Now $$f(a-x)=f(g(x))=g(x)+g(g(x))=g(x)+x=f(x)$$.

In our case define $$g(x)=\frac{3x}{\sqrt{x^2 - 9}}.$$ Indeed, $g$ is an involution as $$g(g(x))=\dfrac{3\frac{3x}{\sqrt{x^2 - 9}}}{\sqrt{\frac{9x^2}{x^2 - 9} - 9}}= \dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{x^2}{x^2 - 9} - 1}}= \dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{9}{x^2 - 9}}}=x. $$

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Like Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$

Clearly $x>0$

Let $\dfrac{3x}{\sqrt{x^2-9}}=y$ so that $x+y=\dfrac{35}4$

$$\implies9x^2=x^2y^2-9y^2\iff\dfrac19=\dfrac1{x^2}+\dfrac1{y^2}=\dfrac{(x+y)^2-2xy}{(xy)^2}$$

Use $x+y=\dfrac{35}4$ and $xy>0$ for find $xy=\dfrac{75}4$

So, $x,y$ are the roots of $$t^2-\dfrac{35}4t+\dfrac{75}4=0$$

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