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How should I approach this problem: $$ 5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2} $$ I've tried squaring both sides but to get rid of all the radicals requires turning it into a quartic equation, which I don't know how to solve? Any tips? Thanks!

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HINT:

Method$\#1:$

Let $2y=\arccos x\implies x=\cos2y,\sqrt{1-x^2}=+|\sin2y|$

As for real $a,\sqrt a\ge0,$

Using the definition of Principal values, $0\le2y\le\pi\implies\sqrt{1-x^2}=+\sin2y$

Again, $0\le2y\le\pi\iff0\le y\le\dfrac\pi2\implies\sin y,\cos y\ge0$

$\implies\sqrt{1-x}=+\sqrt2\sin y,\sqrt{1+x}=+\sqrt2\cos y$

Method$\#2:$

Let $x=\cos2y$

$\sqrt{1-x^2}=|\sin2y|$

$1-x=2\sin^2y\implies\sqrt{1-x}=\sqrt2|\sin y|$ and simialrly $\sqrt{1+x}=\sqrt2|\cos y|$

Now for real $a, |a|=+a$ if $a\ge0,$ else $-a$

Case $\#1:$

$\sin y,\cos y\ge0\implies\sin2y=2\sin y\cos y\ge0$

Case $\#2:$

$\sin y<0,\cos y<0\implies\sin2y=\cdots>0$

Case $\#3:$

$\sin y>0,\cos y<0\implies\sin2y=\cdots$

Case $\#4:$

$\sin y<0,\cos y>0\implies\sin2y=\cdots$

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  • $\begingroup$ Thanks! Your solution is very nice. I never thought of using trigonometry! $\endgroup$ – Matthew Ranouli May 25 '15 at 18:36
  • $\begingroup$ @MatthewRanouli, Wish there is one solution truly algebraic $\endgroup$ – lab bhattacharjee May 25 '15 at 18:51
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    $\begingroup$ There are two real solutions! $\endgroup$ – Dr. Sonnhard Graubner May 25 '15 at 18:54
  • $\begingroup$ @Dr.SonnhardGraubner, $$\cos\left[2y-\arccos\dfrac35\right]=\cos\left[y-45^\circ\right]$$ $$\implies2y-\arccos\dfrac35=2m\pi\pm\left[y-45^\circ\right]$$ $\endgroup$ – lab bhattacharjee May 25 '15 at 18:59
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after squaring two times and sorting and factorizing i got this here $$-4 (25 x-24) \left(100 x^3-75 x+24\right)=0$$

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    $\begingroup$ Although incomplete, this answer has the merit of showing that there are four real solutions: $x=\frac{24}{25}$, a solution just above $-1$, and two around $0.4$. The cosine-substitution method is the standard way to solve a cubic equation with three real solutions. Perhaps Dr. Graubner would care to finish off the answer. $\endgroup$ – John Bentin May 26 '15 at 12:41
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Here is a really nice method: when looking at terms such as $\sqrt{1-x}$, $\sqrt{1+x}$, and $\sqrt{1-x^2}$, think trigonometry!

Make the trig substitution: $$x = \cos{\theta}$$

Now: $$\sqrt{1-x}=\sqrt{2}\sin{\frac{\theta}{2}}$$ $$\sqrt{1+x}=\sqrt{2}\cos{\frac{\theta}{2}}$$ $$\sqrt{1-x^2}=\sin{\theta}$$

The equation with $\theta$ is then:

$$5\left(\sin{\frac{\theta}{2}} + \cos{\frac{\theta}{2}}\right) = 6\cos{\theta} + 8\sin{\theta} $$

Can you take it from here?

Edit: Another trick is used to finish the problem. One of the advantages of trig substitutions is that there are many identities to use. In this case:

$$6\cos{\theta} + 8\sin{\theta} = 10\left(\frac{3}{5}\cos{\theta} + \frac{4}{5}\sin{\theta}\right)$$

Now, if we let $\phi = \sin^{-1}\left(\frac{3}{5}\right)$:

$$6\cos{\theta} + 8\sin{\theta} = 10\left(\sin{\phi}\cos{\theta} + \cos{\phi}\sin{\theta}\right)=10\sin(\phi+\theta)$$

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  • $\begingroup$ Thanks! I never thought of using trigonometry. This was really nicely explained; I figured it out! The answer is $\frac{24}{25}$. $\endgroup$ – Matthew Ranouli May 25 '15 at 18:36

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