Question

Find a Quadratic Equation Corresponding to the Solutions Given Below.
A) x=2, x=-9/2
B) x=-3+2i

Answer 1

if x = 2 and x = -9/2 are the roots then that means

x + 2 and x + 9/2 are the factors

so i can rewrite this as (x+ 2)(2x+9)

expand it to get your quadratic equation

dpes that help?

Answer 2

does*

Answer 3

hmmm

Answer 4

yes, but then how i would i do B?

Answer 5

think it should be \((x- 2)(2x+9)\) to give one zero at \(x=2\)

Answer 6

if one zero is \(-3+2i\) the the other must be its conjugate \(-3-2i\)

Answer 7

so your unenviable job is to multiply this out
\[(x-(-3+2i))(x-(-3-2i))\]
believe it or not, it is not nearly as hard as it looks

Answer 8

oh yes sorry...thanks @satellite73 ...i got careless there

Answer 9

in fact, it is actually very easy, you can just about do it in your head

Answer 10

i can walk you through it if you like

Answer 11

I'm still a little confused on how you would plug it in so it's basically just a conjugate?

Answer 12

yes please walk me through it

Answer 13

yes
first of all if \(a+bi\) is a zero of a real polynomial, then so is \(a-bi\) the conjugate

Answer 14

now we know that it must factor as \((x-(a+bi))(x-(a-bi))\) by the factor theorem
if you have a zero, you can factor
now it is our job to multiply this mess out
but as i said it is not that hard
would you like to do it with the general form
\[ (x-(a+bi))(x-(a-bi))\] or the specific one
\[(x-(-3+2i))(x-(-3-2i))\]?

Answer 15

the specific one please

Answer 16

lets to it with the specific one first
don't distribute, just use "first outer inner last"

Answer 17

first is obviously \(x^2\)
lets hold on to the "outer inner" for a second
last is \((-3+2i)(-3-2i)\) and it is always true that \((a+bi)(a-bi)=a^2+b^2\) a real number
so don't think about minus signs or \(i\) or anything, it is just
\[(-3+2i)(-3-2i)=3^2+2^2=13\]

Answer 18

I see and i get it now. the answer is (-3+2i)(-3-2i)? because of it having a zero so then it has to have its conjugate which is a negative and plus

Answer 19

now for the middle part, outer and inner
you are going to get \((3-2i)x+(3+2i)x\) and when you add them the \(2i\) part goes and you are simply left with \(6x\)

Answer 20

yes the zeros come in conjugate pair
so the original equation must have been
\[x^2+6x+13\]

Answer 21

or rather
\[x^2+6x+13=0\]
if you solve, you will see that the zeros are \(-3\pm2i\)

Answer 22

ahhh i see now. you get that conjugate pair then multiply them both together but then also with the X's to get your quadratic equation.

Answer 23

thanks a lot, i get it all now.

Answer 24

yes you need them both
now lets do it with \(a+bi\) and \(a-bi\) so you can do it quickly if you need it on a test

Answer 25

okay

Answer 26

the quadratic it came from must have been
\[x^2-2ax+(a^2+b^2)=0\]

Answer 27

in your example \(a=-3,b=2\) so you got
\[x^2-2(-3)x+3^2+2^2=0\] or
\[x^2+6x+13=0\]

Answer 28

impress your teacher and do it in 5 seconds

Answer 29

zeros are \(4+3i\) and \(4-3i\) equation must have been
\[x^2-8x+25=0\]

Answer 30

hope it is clear (more or less, may need some practice)

Answer 31

do you have to foil them both once you get the roots?

Answer 32

not sure what you are asking
the way this worked, we were given the roots and we had to come up with the equation
so yes we sort of "foiled" but we get the answer right away

Answer 33

okay i see now given the roots so we could find the quadratic equation

Answer 34

suppose i tell you the zeros are \(2+\sqrt{5}i\) and \(2-\sqrt{5}i\) and ask for the equation they came from
no agony here, i say
\[x^2-2\times 2x+2^2+\sqrt{5}^2=0\] or
\[x^2-4x+9=0\]

Answer 35

the reason i mentioned \(\sqrt{5}\) is because frequently you get a radical when you solve a quadratic equation

Answer 36

okay i see and get everything now, thanks satellite!