Los op:
Oplossing:
Uitwerking
Geeft:
$
\eqalign{
& (x - 1)^2 + \left( {\frac{1}
{3}x^2 } \right)^2 = 1 \cr
& x{}^2 - 2x + 1 + \frac{1}
{9}x^4 = 1 \cr
& x{}^2 - 2x + \frac{1}
{9}x^4 = 0 \cr
& 9x{}^2 - 18x + x^4 = 0 \cr
& x(x^3 + 9x - 18) = 0 \cr
& x = 0 \vee x^3 + 9x = 18 \cr}
$
$x\approx1,6$