Question

Please help! Find a quadratic equation with roots -1+4i and -1-4i. Show your work.

Answer 1

do you know how to factor a quadratic equation? say \(\bf x^2+2x+1\quad ?\)

Answer 2

well \(\bf y=x^2+2x+1\quad ?\)

Answer 3

would that be (x+1)(x+1)?

Answer 4

yes :)

so what that means is that the roots of \(\bf x^2+2x+1 \implies (x+1)(x+1)\)

so that means that \(\bf ((x+1)(x+1) \implies x^2+2x+1\)

so you see, you'd get the polynomial BACK if you simply multiply the roots

Answer 5

so multiply (-1+4i)(-1-4i)?

Answer 6

yes \(\bf x = -1+4i \implies x+1-4i = 0 \implies (x+1-4i)=0\\ \quad \\
x = -1-4i\implies x+1+4i=0 \implies (x+1+4i)=0\\ \quad \\
\textit{so the roots are}\quad (x+1-4i)\quad and\quad (x+1+4i)\qquad so\\ \quad \\

(x+1-4i) (x+1+4i)\\ \quad \\
\textit{recall that }\qquad \color{red}{(a-b)(a+b)} = a^2-b^2\qquad thus\\ \quad \\
(x+1-4i) (x+1+4i)\implies [\color{red}{(x+1)-4i}] [\color{red}{(x+1)+(4i)}]\\ \quad \\ (x+1)^2-(4i)^2
\)

Answer 7

and then you can expand that to get the polynomial, don't forget that \(\bf i^2 = \sqrt{(-1)^2}\implies -1\qquad thus\\ \quad \\
(4i)^2\implies 4^2i^2\implies 16(-1)
\)

Answer 8

okay that makes sense now thank you!

Answer 9

yw