Question

determine the real roots : (5x^2+20) (3x^2-48)

Answer 1

since the numerator is what will cause y=0

you want the numerator to = 0

\[5x^2+20=0\]
\[5x^2=-20\]
\[x^2=-4\]
\[x=\sqrt{-4}\]

\[x=\pm 2i\]

Answer 2

there is only complex roots

Answer 3

thaank you (:

Answer 4

I don't think that's the question.

Answer 5

the answer is + 4 or -4

Answer 6

y =(5x^2+20)(3x^2-48)

has real roots at x = +/- 4

Answer 7

how did you solve for that? O:

Answer 8

roots when y = 0
ie when
(x^2+4) = 0

or

(x^2 - 16) = 0

Answer 9

oh i saw it as division... -.-

Answer 10

yes the second part would be

\[3(x^2-16)\]

Answer 11

\[=(x+4)(x-4)\]

Answer 12

x^2 = -4 x= sqrt(-2) x= +/- 2i *imaginary root*

x^2 =16 x =sqrt(4) = +/- 4

Answer 13

\[(5x^2+20)(3x^2-48)\]

\[5(x^2+4)3(x^2-16)\]
\[15(x^2+4)(x^2-16)\]

\[15 \neq 0\]

\[x^2+4=0\]

\[x^2=-4\]
\[x=\pm 2i\]
\[x^2-16=0\]
\[x^2=16\]
\[x=\pm 4\]

\[15(x-2i)(x+2i)(x-4)(x+4)\]

Answer 14

what does the i stand for ?

Answer 15

nvm i get it (: thank you