# Is there a better way to solve this equation?

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?

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

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

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

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

• $$\dfrac3{\cos(2\cdot \arctan\left(\frac13\right)}=5$$
• $$\dfrac3{\cos(2\cdot \arctan\left(\frac12\right)}=\frac{15}4$$

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 $$g(g(x))=\dfrac{3\frac{3x}{\sqrt{x^2 - 9}}}{\sqrt{\frac{9x^2}{x^2 - 9} - 9}}= \dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{x^2}{x^2 - 9} - 1}}= \dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{9}{x^2 - 9}}}=x.$$

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 $$t^2-\dfrac{35}4t+\dfrac{75}4=0$$