Question

HELP

Solve the equation
16x^2+9=0

a. 3/4
b. 3/4i
c. 4/3i
d. 9/16i

Answer 1

rearrange to make the equation equal to a constant
\[16x^{2}+9=0\]
\[16x^{2}=-9\]
solve for x

\[x^{2}=\frac{ -9 }{ 16 }\]
\[x=\sqrt{(\frac{ -9 }{ 16 })} \]

\[\sqrt{\frac{ a }{ b }}=\frac{ \sqrt{a} }{ \sqrt{b} }\]
\[x=\frac{ \sqrt{-9} }{ \sqrt{16} }\]

\[\sqrt{-a}=\sqrt{-1}\sqrt{a}\]
\[\sqrt{-1}= i\]
\[x=\frac{ \sqrt{-1}\sqrt{9} }{ \sqrt{16} }\]

Answer 2

so its d? @amorfide

Answer 3

no I didn't finish it, I left it for you to finish

you know sqrt9 is 3
sqrt -1 is i
sqrt 16 is 4

simplify what I gave you and you get your answer

Answer 4

3/4i

Answer 5

well done

Answer 6

thanks :)