Question

Find all the zeros for the equation
-3x^2+27x^2+1200=0

Answer 1

-3x^2 + 27x^2 + 1200 = 0

Start by combining the x^2 terms.
24x^2 + 1200 = 0

Then divide everything by 24...

Answer 2

@vivianrush are you working on this one still?

Answer 3

i divide 1200 by 24 and got 50

Answer 4

You also divide the 24x^2 term by 24 to get just x^2.

It should look like:
(24/24)x^2 + 1200/24 = 0

x^2 + 50 = 0

Answer 5

so then its x^2 +50=0

im suppose to find all the zeros of the equation

Answer 6

Can they be imaginary number roots? Or are you only allowed to have real number roots in this class?

Answer 7

real number roots

Answer 8

x^2 + 50 = 0

Subtract 50 from both sides:

x^2 + 50 - 50 = 0 - 50

x^2 = -50

Then you would need to take the square root of both sides... ordinarily you would get a "plus or minus something" as your two zeroes... but in this case, the stuff under the square root is a negative 50... you can't take the square root of a negative number and get real number roots (or "zeroes").

Answer 9

thank you so much. i understand this a lot better now

Answer 10

If it had been like x^2 = 25 or whatever, then you would take square root and get zeroes of x equal to plus and minus 5 (or "5 and -5").

But if the stuff under the square root is negative, the answer is "no real zeroes" or something similar.

Answer 11

Glad to help :)