What is the question attached asking me to do?
You have to prove that the given polynomial will yield no solutions in \(\mathbb{C}\).
what is C?
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.
what part of what you said refers to the independent of part?
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.
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
Then I thought that a=qx^2, b=3x-6 and c=4q
firs of all, your b can't have the -6 in it that has to be a part of the constant, c
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
So do I first solve for q?
by "a variable" I meant that a, b, and c do not include the x in this case, they will include q though
no you do not "solve" for q first identify a, b, and c in your expression of the form\[ax^2+bx+c\]
a=1 b=3 and c=-6?
or maybe a=q?
yes a=q
now what about the others ? :)
and b=3 is right as well
c=-6?
-4q
no, c has to be \(everything\) that is left over so both of the above together are c
so c=-6-4q?
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?
b^2-4ac
yes! and what is that in your case?
9-4(q)(-6-4q) is that what I have to solve?
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
maybe add 4q to the inequality?
just simplify what you had for the discriminant and tell me what you get
9-4(q)(-6-4q) 9-4q(-6)-4(4q)=???
9-4(-6q-4q^2)
distribute the 4 to get rid of the parentheses
the -4 I mean
9-4(q)(-6-4q) 9-4q(-6)-4(-4q)=???
9+24q+16q^2
wait.
16q^2+24q+9???
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...
right, I put that back into the b^2-4ac and it =0
?
also I completed the square and got -3/4 for both soultions
that is redundant; it comes from the fact that we know this is the discriminant of our problem
there is no solving to be done here just factor\[16q^2+24q+9\]
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
ok 16*9=144 so (x+12)^2??
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
the point to me is that the discriminant of that equation is always a perfect square, so it is always \(\ge0\)
hence the roots are always real
wait, but I factored it. cant we continue???
you didn't factor \[16q^2+24q+9\]correctly
oh what is it then?
I just wrote it above
\[(16q^2+24q+9=(4q+3)(4q+3)=(4q+3)^2\ge0\]because a number squared is always \(\ge0\)
so the discriminant is always \(\ge0\)
which means the roots are always real
that was fun. thanks
yeah it was welcome :)
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