Question

\[\frac{ q^2 }{ 4 } - \frac{ 9q^2 }{ 20 } = 18\], how do i convert this to standard form so i can find the intercepts? This is confusing!

Answer 1

draw my sausage!

Answer 2

\[\frac{ q^2 }{ 4 } - \frac{ 9q^2 }{ 20 } = 18\], how do i convert this to standard form so i can find the intercepts? This is confusing!
@Catlover5925

Answer 3

First multiply the whole thing by 20 to get rid of the fractions.

Answer 4

\[
5q^2-9q^2=360
\]

Answer 5

here let @wio help he is better at explaining then me

Answer 6

Do you think you can do the rest @anom123 ?

Answer 7

Does that mean b and c is 0?

Answer 8

Wait, what did you mean by intercepts?

Answer 9

ah i just think i solve for q2 in this one

Answer 10

Yeah, so do you think you can solve for it?

Answer 11

just confused about one thing, could we have multipled by 4 instead? and when we divided by 20, how come the 4 wasn't touched?

Answer 12

You see \(20/4 = 5\). That is what happened.

Answer 13

and how come 9q wasn't multipled by 20? fractions always confuse me :/

Answer 14

\[20\times \frac{ q^2 }{ 4 } - 20\times\frac{ 9q^2 }{ 20 } = 20\times18\]

Answer 15

\[5q^2- 9q^2 = 360\]

Answer 16

THANKS, that made alot of sense, let me solve the probelm now

Answer 17

would it be square root of 4?

Answer 18

There are no real solutions. Only imaginary solutions.

Answer 19

\[
q = \pm \sqrt{-90} = \pm 3i\sqrt{10}
\]

Answer 20

can you show the steps you took to get from sqrt -90 to +- 3i sqrt 10?
i understand now that there is no imaginary solution because if it's a negative radicand then it is imaginary

Answer 21

\[
5q^2- 9q^2 = 360\\
-4q^2=360\\
q^2 = -\frac{360}{4} = -90

\]

Answer 22

\[
\sqrt{-90} = \sqrt{(-1)(2)(3^2)(5)} = 3i\sqrt{(2)(5)} = 3i\sqrt{10}
\]

Answer 23

alright, i think i understand everything now, thank you very much wio!!