Question

how do you find the solution set of x^2+25=0?? i know the answer is {5i,-5i} but not sure why

Answer 1

subtract \(25\) from both sides, then take the square root

Answer 2

i did but i dont know why the imaginary numbers are included in the answer

Answer 3

you get \(x^2=-25\) so \(x=\pm\sqrt{-25}\)

Answer 4

all you are doing is rewriting \(\sqrt{-25}+\sqrt{25\times (-1)}=\sqrt{25}\sqrt{-1}=5\sqrt{-1}\)

Answer 5

and then rewriting \(\sqrt{-1}\) as \(i\) which is why it is \(5i\)

Answer 6

i understand it now! thank you so much! :D

Answer 7

yw

Answer 8

wait... theres 5i but how do get -5i??

Answer 9

because of the + or -?

Answer 10

oh because it is exactly the same is if you had solve \(x^2-25=0\)
you get \(x^2=25\) and so \(x=5\) or \(x=-5\) always two solutions

Answer 11

in this case you get \(x^2=-25\) so \(x=5i\) or \(x=-5i\) as above

Answer 12

like -5x5=25 and 5x5 =25?

Answer 13

*-5x-5=25

Answer 14

yes exactly

Answer 15

in this case \((-5i)(-5i)=25i^2=-25\)

Answer 16

thanks again :)