Respuesta :
Answer:
1,-1 are rational numbers
Step-by-step explanation:
Given equation: [tex]f(x) = x^6-2x^4-5x^2+6[/tex]
Roots of f(x) are : [tex]x^6-2x^4-5x^2+6=0[/tex]
⇒ [tex](x-1)(x+1)(x^2-3)(x^2+2)[/tex]
⇒ [tex](x-1)(x+1)(x+\sqrt{3})(x-\sqrt{3})(x+\sqrt{2})(x-\sqrt{3})[/tex]
⇒[tex](x=1)(x=-1)(x=\sqrt{3})(x=-\sqrt{3})(x=\sqrt{2})(x=-\sqrt{2})[/tex]
⇒[tex]x=1,-1,\sqrt{3},-\sqrt{3},\sqrt{2},-\sqrt{2}[/tex]
from these roots only -1,1 is consider as rational numbers as all integers are rational numbers.
Rest [tex]x=\sqrt{3},-\sqrt{3},\sqrt{2},-\sqrt{2}[/tex] are irrational numbers.
Answer:
B
Step-by-step explanation:
the roots would be 1 and -1 when it comes to the graph
