Question

What are the solutions?
16x^2+9=0

Answer 1

they are imaginary

\(x^2= - \dfrac{9}{16}\)

on an argand diagram you will find 2 solutions

Answer 2

@IrishBoy123 Is the argand diagram simply a graph or is it something else? I've never seen it or heard of it before..

Answer 3

your solution is

\[x= \sqrt{ - \dfrac{9}{16}}\]

i don't know what you're learning but taking the sqrt of a -ve number is a bit of a challenge. you'll get 2 answers.

is this what you expected ??

complex numbers have 2 squares roots, 3 cube roots, 4 quartic roots, all symmetrically arranged around the complex plane origin.

if this is jibberish to you, you should say. perhaps there is a typo in what we are talking about :(

Answer 4

@IrishBoy123 Hm idk, this is a screen shot of the questions.

Answer 5

And these are the options, and like it says, the first one is correct (I took a quiz and I did bad on it so I'm reviewing the questions I messed up on on OpenStudy so I can be prepared to answer questions like them in the future lol)

Answer 6

\[x^2=-\frac{ 9 }{ 16 }=\frac{ 9 }{ 16 }\times -1=\frac{ 9 }{ 16 }\iota^2\]
\[x^2-\frac{ 9 }{ 16 }\iota^2=0\]
\[x^2-\left( \frac{ 3 }{ 4 }\iota \right)^2=0\]
\[\left( x+\frac{ 3 }{ 4 }\iota \right)\left( x-\frac{ 3 }{ 4 } \iota \right)=0\]
x=?

Answer 7

@sshayer 3/4i, -3/4i?

Answer 8

correct.

Answer 9

@sshayer Thank you for your help :)