Question

Solve the following system:
x^2+y^2=9
x^2-y^2=1

The solution set is {( sq. rt of____,____),(sq.rt of____,-_____),(-sq.rt of____,_____),(-sq. rt of___,-_____)}

Answer 1

If you add vertically, you get 2x^2=10. So \[x= \pm \sqrt{5}\] Then you solve for y in the first equation, and you get y^2=-16. So \[x= \pm 4i\] So your answers are:\[(\pm \sqrt{5} , \pm4i)\]

Answer 2

but why i?

Answer 3

because you are taking the square root of a negative number. \[i=\sqrt{-1}\]

Answer 4

hmmmm

Answer 5

You may eliminate by adding both equations

Answer 6

lets try this one again

Answer 7

\( \color{Black}{\Rightarrow 2x^2 = 10 }\)
\( \color{Black}{\Rightarrow x^2 = 5 }\)
\( \color{Black}{\Rightarrow x = \sqrt5 }\)

Now substitute \(x\) back to any of the equations and find what \(y\) is.

Answer 8

\( \color{Black}{\Rightarrow 5 + y^2 = 9 }\)
\( \color{Black}{\Rightarrow y^2 = 4 }\)

\( \color{Black}{\Rightarrow \Huge y = 2 }\)

Where did that \(i\) come from? o.O

Answer 9

add both equations together, you get
\[2x^2=10\]
\[x^2=5\]
\[x=\sqrt{5}\] or
\[x=-\sqrt{5}\]

Answer 10

I always forget the \(\pm \) part ugh

Answer 11

that's better! a nice real answer

Answer 12

yeah so there are 4 answer right?

Answer 13

lol

Answer 14

i like to crack jokes up here. hmm.

Answer 15

oops...my bad.

Answer 16

\[\sqrt{5},2\]
\[\sqrt{5},-2\]
\[-\sqrt{5},2\]
\[-\sqrt{5},-2\]

Answer 17

That's like all the possible permutations of the + and - signs :P

Answer 18

y=2.03, -2.03

Answer 19

to be exact

Answer 20

thanks guys :D