Respuesta :

apologiabiology
using trial and error
x^4+5x^2-36 factors to (x-2)(x+2)(x^2+9)
2x^2+9x-5 factors to (x+5)(2x-1)
so

(x-2)(x+2)(x^2+9)(x+5)(2x-1)=0

set each to zero
x-2=0
x=2

x+2=0
x=-2

x^2+9=0
x^2=-9
x=3√-1
x=3i

x+5=0
x=-5

2x-1=0
2x=1
x=1/2=0.5



roots are
x=-5-2,0.5,2, and the imaginary root 3i
caylus
Hello,

1)
X^4+5x²-36=0
Δ=5²+4*36=169=13²
x²=(-5-13)/2=-9 or x²=(-5+13)/2=4
x^4+5x²-36=(x²+9)(x²-4)=(x-2)(x+2)(x²+9)

2)
2x²+9x-5=0
Δ=9²+4*5*2=121=11²
x=(-9-11)/4=-5 or x=(-9+11)/4=1/2

2x²+9x-5=(x+5)(2x-1)

(x^4+5x^2-36)(2x^2+9x-5)=(x+5)(2x-1)(x+2)(x-2)(x²+9)

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