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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0easier than you think

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \(3+2i\) is a zero so is \(32i\) the quadratic factors as \[(x(3+2i))(x(32i))\] when you multiply out you will get \(x^2\) as the first term last term \(n\) will be \(3+2i)(32i)=3^2+2^2=14\)and the "middle term" will be\[(3+2i)x(32i)x=6x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually \(3^2+2^2=13\) but whatever

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the original quadratic was \[x^26x+13\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0once you see this you can know if \(a+bi\) is a zero, s is \(abi\) and the original quadratic is \[x^22ax+a^2+b^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0btw you can always work backwards: \[x=3+2i\] \[x3=2i\] \[(x3)^2=(2i)^2=4\] \[(x3)^2+4=0\] \[x^26x+9+4=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so how do I get the value of m from there?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^2+mx+n=0\] \[x^26x+13=0\] what is \(m\)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the original question asked for the value of m

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large(3+2i)^2+m(3+2i)+n=0\] \[\Large 9+12i+4i^2+3m+2 mi+n=0\] \[\Large (3m+n)+ 2mi=12i5\] \[\Large 3m+n=5 \\2m=12\\m=6\\n=13\]
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