Question

Help with this question- inside

Answer 1

What is

\[\huge\ \sqrt{-1}=..\]

Answer 2

@liizzyliizz

Answer 3

well, you have to know that \[\sqrt{-1}=i\].. then you can sum the equals terms at the denominator: \[\frac{ \sqrt{-49} }{ 7-2i-4+9i }=\frac{ \sqrt{-49} }{ 3+7i }\] later, you can express your equation like:\[\frac{ \sqrt{-49} }{ 3+7i }=\frac{ \sqrt{-49} }{ 3+7\sqrt{-1}}\] and you can introduce the number 7 in the square so:\[\frac{ \sqrt{-49} }{ 3+\sqrt{49\times-1}}=\frac{ \sqrt{-49} }{ 3+\sqrt{-49} }\] and for the factorization you know that \[(a+b)\times(a-b)=a ^{2}-b ^{2}\] then we multiplicate the equation for the factorization : \[\frac{ \sqrt{-49} }{ 3+\sqrt{-49}}\times\frac{ {3-\sqrt{-49}} }{ 3-\sqrt{-49}}\] and resolving \[\frac{ (3+\sqrt{-49})\times \sqrt{-49} }{ 3^{2} - 49 }\].. at the end: \[\frac{ 49+21i }{ 5 }\]