which describes the number and type of roots of the equation x^3+121x=0?

Question

a. 1 real root, two imaginary roots.
b. 2 real roots, 1 imaginary root.
c. 3 real roots
d. 3 imaginary roots

Answer 1

after working it out, i got A

Answer 2

Because it's = 0 then all the roots are real?

Although, I could also argue that the whole thing is imaginary since theoretical mathematics is really just a mental phenomenon. There could be real phenomena that could be accounted for by mathematics, but the numbers are only because we imagine them, such as they are. I mean, would our numbering system not be different if we had 12 fingers and 12 toes? And wouldn't there still be people bragging about how perfect it was because number sets of 12 were "perfect." Anyhow, that's all in our heads. All numbers are imaginary.

Anyhow, there should be more hot chicks posing math questions in this world... Maybe that's what this country really needs to solve our math problems (social and perceived, although possibly not real).

Interesting... Now that I see everyone else's answer's I'm tempted to change, but wouldn't it be the same if x=0? So all roots = 0, hence, it is all the specific "real" number 0. Otherwise, I'm starting to think that maybe Math tries too hard to impress itself with itself... which is why I got bored quickly and didn't accept the scholarship offer... That and I didn't see any women doing math... I guess you could say it was an opportunity cost decision... ;)

Answer 3

(a) 1 real root, two imaginary roots

x^3 + 121 x = x ( x^2 + 121 ) = 0

x = 0 ----> Real Root

x^2 + 121 = 0

x^ 2 = -121

x = - squareroot( -121) and x = + squareroot( -121)

Since 121 = 11^2,

x = - 11 i and x = + 11 i -----> Imaginary Roots

Good luck

Answer 4

1 real and 2 imaginary.

0 is the real root and

the imaginary roots are plus and minus 11i.

factor out an x of each term then use quadratic formula for solutions or simply use the discriminant (b^2 - 4ac = -484) to see that there are two complex or imaginary roots.

Answer 5

x(x^2 + 121) = 0
x = 0, -11i, 11i

so a. 1 real root, two imaginary roots

Answer 6

That equation factors into x(x^2+121)=0

x=0 creates one real root.
x^2+121=0 becomes x^2=-121 and produces two imaginary roots.

a. is the answer.

Answer 7

factor it an find out
x^3+121x=0
x(x^2+121)=0
x=0 or x^2+121=0
x=0 or x^2=-121
x=0 or x=+-sqrt(-121)
x=0 or x=+-11i
so 1 real root...x=0
and two imaginary ...-11i and +11i

Answer 8

x^3 + 121x = 0

x(x^2 + 121) = 0

x = 0

x^2 + 121 = 0
x^2 = -121
x = +/- 11i

One real root, two imaginary.

Answer 9

x^3+121x=0
x(x^2+121)=0
x=0
x^2+121=0
x^2=-121
x=+-sqr(121)
x=+-11i

so x=0, x=11i, x=-11i

Choice a
x=0 is real, and x=11i, x=-11i are imaginary