Question

x⁴-5x +4=0 me ajudem com essa equação biquadrada

Answer 1

x² = y, portanto x$x^{4}$ = y²

y² - 5y + 4 = 0
y1= 4
y2= 1

x² = y     p/ y1= 4
$\sqrt{x²}$ = $\sqrt{4}$
x1= 2
x2= -2


x² = y     p/ y2 = 1
$\sqrt{x²}$ = $\sqrt{1}$
x3= 1
x4= -1

S= {-2,-1,1,2}