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Solving the equation of the form

$$1-3x^2+3x\sqrt{1-2x^2}=0 $$

Is cumbersome since setting $t=1-2x^2$ does not yield an explicit quadratic formula in terms of t. There is some trix to this, but I already tried to solve it looking at $\sqrt{1-2x^2} $ as the $b$ in $ax^2+bx+c=0$ and also to insert $t$ as above and that yields nothing good. How does one go around this?

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HINT:

We have $$3x\sqrt{1-2x^2}=3x^2-1$$

Square both sides to form a Quadratic Equation in $x^2$

Solve it & identify the extraneous roots introduced due to squaring

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  • $\begingroup$ @Raxel, Please proceed & let me know in case you are stuck $\endgroup$ – lab bhattacharjee Oct 9 '14 at 4:28
  • $\begingroup$ So we get in fact 4 solutions, but since we squared in the beginning we need to get rid of two of the four. Is there any better way to know which ones work other than plugging them into the initial equation? $\endgroup$ – Raxel Oct 9 '14 at 4:37
  • $\begingroup$ @Raxel, As for real $x,1-2x^2\ge0, x^2\le\dfrac12$ $\endgroup$ – lab bhattacharjee Oct 9 '14 at 4:47
  • $\begingroup$ Yeah that's true but all solutions will be s.t. $0 \leq x^2 \leq \frac{1}{2} $, so I will just plug them in. $\endgroup$ – Raxel Oct 9 '14 at 4:56
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Hint:

Another way is to set $$ \sin \theta = \sqrt 2 \ x$$ and then solve an equation of the form:

$$ a\cos \phi + b \sin \phi = c $$

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  • 1
    $\begingroup$ Was thinking to add this, then rejected suspecting overkill or not? $\endgroup$ – lab bhattacharjee Oct 9 '14 at 4:30
  • $\begingroup$ @lab bhattacharjee: Yeah definitely counts as overkill. Squaring is much easier. Just added it as an alternate method. $\endgroup$ – Ishfaaq Oct 9 '14 at 4:35
  • $\begingroup$ Its always nice to share alternative solutions.Please do continue $\endgroup$ – lab bhattacharjee Oct 9 '14 at 4:49

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