Question

If 3+2i is a solution for x^2+mx+n , where m and n are real numbers, what is the value of m?

Answer 1

easier than you think

Answer 2

if \(3+2i\) is a zero so is \(3-2i\) the quadratic factors as \[(x-(3+2i))(x-(3-2i))\]
when you multiply out you will get \(x^2\) as the first term
last term \(n\) will be \(3+2i)(3-2i)=3^2+2^2=14\)and the "middle term" will be\[-(3+2i)x-(3-2i)x=-6x\]

Answer 3

actually \(3^2+2^2=13\) but whatever

Answer 4

the original quadratic was
\[x^2-6x+13\]

Answer 5

m should be -6

Answer 6

once you see this you can know if \(a+bi\) is a zero, s is \(a-bi\) and the original quadratic is
\[x^2-2ax+a^2+b^2\]

Answer 7

btw you can always work backwards:
\[x=3+2i\]
\[x-3=-2i\]
\[(x-3)^2=(-2i)^2=-4\]
\[(x-3)^2+4=0\]
\[x^2-6x+9+4=0\]

Answer 8

okay so how do I get the value of m from there?

Answer 9

\[x^2+mx+n=0\]
\[x^2-6x+13=0\] what is \(m\)?

Answer 10

the original question asked for the value of m

Answer 11

so m is -6?

Answer 12

Thanks!!

Answer 13

\[\Large(3+2i)^2+m(3+2i)+n=0\]
\[\Large 9+12i+4i^2+3m+2 mi+n=0\]
\[\Large (3m+n)+ 2mi=-12i-5\]
\[\Large 3m+n=-5 \\2m=-12\\m=-6\\n=13\]