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Can we have a triangle with sides $1$, $x$ and $x^2$? And what could $x$ be?

I try to approach this question by making 3 inequalities.
$1+x>x^2$,
$1+x^2>x$,
$x^2+x>1$

and they come with different quadratic inequalities
$x^2-x-1<0$ (solution is $(1+\sqrt 5)/2 > x > (1-\sqrt 5)/2$ ) ;
$x^2-x+1>0$ (solution is $x > (-1+\sqrt 5)/2$ or $x < (-1-\sqrt 5)/2$ )
$x^2+x-1>0$ (no real solution)

Then I start to struggle with the next step.... Thank you

  • 0
    Struggling with the next step probably has a lot to do with getting the previous step wrong. Nice start, though!2012-12-04

5 Answers 5