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lys-0071 [83]
3 years ago
11

Find a quadratic equation with roots -1, 4i and -1 - 4i

Mathematics
1 answer:
vazorg [7]3 years ago
3 0
\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }i^2=-1\\\\
-------------------------------\\\\
\begin{cases}
x=-1+4i\implies &x+1-4i=0\\
x=-1-4i\implies &x+1+4i=0
\end{cases}
\\\\\\
(x+1-4i)(x+1+4i)=\stackrel{y}{0}
\\\\\\\
[(x+1)~~-~~(4i)][(x+1)~~+~~(4i)]=y
\\\\\\\
[(x+1)^2~~-~~(4i)^2]=0\implies (x^2+2x+1)~-~(4^2i^2)=y
\\\\\\
(x^2+2x+1)~-~(16\cdot -1)=y\implies (x^2+2x+1)~-~(-16)=y
\\\\\\
x^2+2x+1~~+16=y\implies x^2+2x+17=y
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