Question

quais as raízes de x⁴-5x²+4=0​

Answer 1

$y = {x}^{2} \\ {y}^{2} - 5y + 4 = \\ delta = \sqrt{ ({ - 5})^{2} - 4 \times 1 \times 4 } \\ delta = \sqrt{25 - 16} \\ delta = \sqrt{9} \\ delta = 3 \\ y = \frac{ - 5 + - 3}{2 \times 1} = \\ y. = - \frac{5 - 3}{2} = \frac{ - 8}{2} = - 4 \\ y.. = \frac{ - 5 + 3}{2} = \frac{ - 2}{2} = - 1$

logo x=4, x=1, x=-4, x=-1

Answer 2

$\large\boxed{{olaaa}}}}}$

X⁴ — 5x² + 4 = 0

seja : x⁴ = t² e x² = t

logo:

t² — 5t + 4 = 0

onde:

a = 1 ; b = —5 e c = 4

∆ = (—5)² — 4 • 1 • 4

∆ = 25 — 16

∆ = 9

• Bhaskara:
• t¹'² = ( -b±√∆)/2•a
• t¹'²=(5±√9)/2•1
• t¹'2 = (5±3)/2

• t¹ =( 5+3)/2 = 8/2 = 4
• t² = (5—3)/2 = 2/2 = 1

Como a nossa incógnita é x,temos que:

x¹'² = ±√t¹

x¹'² = ±√4 = ±2

x³'⁴ = ±√t²

x³'⁴ = ±√1 = ±1

Sol: { —1 , —2 , 1 , 2 }

Espero ter ajudado bastante!)