What is the question attached asking me to do?

Question

anonymous · Answer

attachment

across · Answer

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

anonymous · Answer

what is C?

across · Answer

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.

anonymous · Answer

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

across · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

turingtest · Answer

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

turingtest · Answer

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

anonymous · Answer

So do I first solve for q?

turingtest · Answer

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

turingtest · Answer

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

anonymous · Answer

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

anonymous · Answer

or maybe a=q?

turingtest · Answer

yes a=q

turingtest · Answer

now what about the others ?
:)

turingtest · Answer

and b=3 is right as well

anonymous · Answer

c=-6?

anonymous · Answer

-4q

turingtest · Answer

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

anonymous · Answer

so c=-6-4q?

turingtest · Answer

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?

anonymous · Answer

b^2-4ac

turingtest · Answer

yes!
and what is that in your case?

anonymous · Answer

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

turingtest · Answer

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

anonymous · Answer

maybe add 4q to the inequality?

turingtest · Answer

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

turingtest · Answer

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

anonymous · Answer

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

turingtest · Answer

distribute the 4 to get rid of the parentheses

turingtest · Answer

the -4 I mean

turingtest · Answer

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

anonymous · Answer

9+24q+16q^2

anonymous · Answer

wait.

anonymous · Answer

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

turingtest · Answer

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...

anonymous · Answer

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

turingtest · Answer

?

anonymous · Answer

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

turingtest · Answer

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

turingtest · Answer

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

phi · Answer

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

anonymous · Answer

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

turingtest · Answer

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

turingtest · Answer

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

turingtest · Answer

hence the roots are always real

anonymous · Answer

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

turingtest · Answer

you didn't factor $$16q^2+24q+9$$correctly

anonymous · Answer

oh what is it then?

turingtest · Answer

I just wrote it above

turingtest · Answer

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

turingtest · Answer

so the discriminant is always $\ge0$

turingtest · Answer

which means the roots are always real

anonymous · Answer

that was fun. thanks

turingtest · Answer

yeah it was
welcome :)