Answer:
+ or - square root of 5, and + or - i
Step-by-step explanation:
(x^2 + 5)(x^2 + 1)= 0
Set each set of parentheses equal to zero:
x^2 + 5 = 0; x^2 = -5, so x will equal plus or minus i times the square root of 5.
X^2 +1 = 0; x^2 = -1, so x will equal plus or minus i times the square root of one, which is just i
Answer: The solutions are i and 5i.
Step-by-step explanation:
Since we have given that
[tex]x^4+6x^2+5=0[/tex]
By using Substitution, we use
[tex]y=x^2[/tex] in the above equation:
[tex]y^2+6y+5=0\\\\y^2+5y+y+5=0\\\\y(y+5)+1(y+5)=0\\\\(y+1)(y+5)=0\\\\y=-1,-5[/tex]
Now, put the value of y in the equation [tex]y=x^2[/tex].
Put y = -1,
[tex]-1=x^2\\\\x=\sqrt{-1}=i[/tex]
Similarly, put y = -5
[tex]-5=x^2\\\\x=\sqrt{-5}=5i[/tex]
Hence, the solutions are i and 5i.
