Question

Help me with these two problems, I don't really understand how to use the "imaginary number" i
1.) x^2=-25
2.) 3s^2=-64
Thanks xx

Answer 1

by definition, \[i^2 = -1\]

Answer 2

so \[x^2=-25 -> x=i5\]

Answer 3

you solve the equation like a normal one, but whenever you encounter \[\sqrt(-1)\], replace it by i

Answer 4

1. \(x^2 = -25\)

Take the square root of both sides, you will end up with 2 values of x.\[x= \pm \sqrt{-25} = \sqrt{-1} \cdot \pm\sqrt{25}\]\[i = \sqrt{-1}\]\[\sqrt{-25} =\pm 5i\]

Answer 5

NOTE: there is one thing you CANNOT do, that is
\(\sqrt{25}=\sqrt{-5*-5}=\sqrt{-5}\sqrt{-5}=i\sqrt{5}*i\sqrt{5}=-1*\sqrt{25}=-5\)

Answer 6

2. \(3s^2 = -64\) divide both sides by 3 to isolate s^2 \[s^2 = -\frac{64}{3}\]take the square root of both sides.\[s= \pm \sqrt{-\frac{64}{3}}\]Pull out \(i = \sqrt{-1}\)\[s= \sqrt{-1} \cdot \pm \sqrt{\frac{64}{3}}\]\[s=\pm i\sqrt{\frac{64}{3}}\]

Answer 7

Are you confused about anything?

Answer 8

No I got it, thank you so much!

Answer 9

Awesome :)