Question

What is the question attached asking me to do?

Answer 1

You have to prove that the given polynomial will yield no solutions in \(\mathbb{C}\).

Answer 2

what is C?

Answer 3

It's the set of complex numbers. Assuming that \(q\in\mathbb{R}\), you have to show that the polynomial has roots in \(\mathbb{R}\) only.

Answer 4

what part of what you said refers to the independent of part?

Answer 5

I don't understand your question. Let me ask you, what have you tried thus far? If you tell me, I'll have a better clue as to your mathematical understanding.

Answer 6

Well I haven't actually tried anything yet, but my 1st thought was to find q, and then maybe the roots of qx^2+3x-6

Answer 7

Then I thought that a=qx^2, b=3x-6 and c=4q

Answer 8

firs of all, your b can't have the -6 in it
that has to be a part of the constant, c

Answer 9

actually none of those should have a variable in them; they are the coefficients\[ax^2+bx+c\]so what is the coefficient of x^2 ? that is a
what is the coefficient of x? that will be your b
what is the left over? that will be c

Answer 10

So do I first solve for q?

Answer 11

by "a variable" I meant that a, b, and c do not include the x
in this case, they will include q though

Answer 12

no you do not "solve" for q
first identify a, b, and c in your expression of the form\[ax^2+bx+c\]

Answer 13

a=1 b=3 and c=-6?

Answer 14

or maybe a=q?

Answer 15

yes a=q

Answer 16

now what about the others ?
:)

Answer 17

and b=3 is right as well

Answer 18

c=-6?

Answer 19

-4q

Answer 20

no, c has to be \(everything\) that is left over
so both of the above together are c

Answer 21

so c=-6-4q?

Answer 22

yes :)
now what formula do we use to find out about the nature of a quadratics roots (whether they are complex, real, or double)?
do you remember?

Answer 23

b^2-4ac

Answer 24

yes!
and what is that in your case?

Answer 25

9-4(q)(-6-4q) is that what I have to solve?

Answer 26

careful how you use the word "solve"
usually we want to know if\[b^2-4ac>0\]\[b^2-4ac<0\]or if\[b^2-4ac=0\]
so we are comparing the (usually just numbers) on the left with the right
but here we have that pesky variable q, so we need a clever way around it
first simplify what you have

Answer 27

maybe add 4q to the inequality?

Answer 28

just simplify what you had for the discriminant and tell me what you get

Answer 29

9-4(q)(-6-4q)
9-4q(-6)-4(4q)=???

Answer 30

9-4(-6q-4q^2)

Answer 31

distribute the 4 to get rid of the parentheses

Answer 32

the -4 I mean

Answer 33

9-4(q)(-6-4q)
9-4q(-6)-4(-4q)=???

Answer 34

9+24q+16q^2

Answer 35

wait.

Answer 36

16q^2+24q+9???

Answer 37

yeah, same thing
now what kind of tricks can we do to this
the only thing I can think is to try to factor it, so why don't you try that and see if that tells us something...

Answer 38

right, I put that back into the b^2-4ac and it =0

Answer 39

?

Answer 40

also I completed the square and got -3/4 for both soultions

Answer 41

that is redundant; it comes from the fact that we know this is the discriminant of our problem

Answer 42

there is no solving to be done here
just factor\[16q^2+24q+9\]

Answer 43

also I completed the square and got -3/4 for both soultions

that means both roots are at the vertex (lowest point) of the parabola. If the lowest point of the parabola is >=0 you have proven that the discriminant (of the original equation) will never by negative, and so the roots will always be real

Answer 44

ok 16*9=144 so (x+12)^2??

Answer 45

ok I suppose that is a convincing argument, I hadn't considered that
what I was going for was\[16q^2+24q+9=(4q+3)^2\ge0\]but if you invoke the idea that this is a parabola I guess that is okay

Answer 46

the point to me is that the discriminant of that equation is always a perfect square, so it is always \(\ge0\)

Answer 47

hence the roots are always real

Answer 48

wait, but I factored it. cant we continue???

Answer 49

you didn't factor \[16q^2+24q+9\]correctly

Answer 50

oh what is it then?

Answer 51

I just wrote it above

Answer 52

\[(16q^2+24q+9=(4q+3)(4q+3)=(4q+3)^2\ge0\]because a number squared is always \(\ge0\)

Answer 53

so the discriminant is always \(\ge0\)

Answer 54

which means the roots are always real

Answer 55

that was fun. thanks

Answer 56

yeah it was
welcome :)