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Mathematics 76 Online
OpenStudy (anonymous):

What is the question attached asking me to do?

OpenStudy (anonymous):

OpenStudy (across):

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

OpenStudy (anonymous):

what is C?

OpenStudy (across):

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.

OpenStudy (anonymous):

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

OpenStudy (across):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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

OpenStudy (turingtest):

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

OpenStudy (anonymous):

So do I first solve for q?

OpenStudy (turingtest):

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

OpenStudy (turingtest):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

or maybe a=q?

OpenStudy (turingtest):

yes a=q

OpenStudy (turingtest):

now what about the others ? :)

OpenStudy (turingtest):

and b=3 is right as well

OpenStudy (anonymous):

c=-6?

OpenStudy (anonymous):

-4q

OpenStudy (turingtest):

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

OpenStudy (anonymous):

so c=-6-4q?

OpenStudy (turingtest):

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?

OpenStudy (anonymous):

b^2-4ac

OpenStudy (turingtest):

yes! and what is that in your case?

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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

OpenStudy (anonymous):

maybe add 4q to the inequality?

OpenStudy (turingtest):

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

OpenStudy (turingtest):

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

OpenStudy (anonymous):

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

OpenStudy (turingtest):

distribute the 4 to get rid of the parentheses

OpenStudy (turingtest):

the -4 I mean

OpenStudy (turingtest):

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

OpenStudy (anonymous):

9+24q+16q^2

OpenStudy (anonymous):

wait.

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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...

OpenStudy (anonymous):

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

OpenStudy (turingtest):

?

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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

OpenStudy (turingtest):

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

OpenStudy (phi):

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

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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

OpenStudy (turingtest):

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

OpenStudy (turingtest):

hence the roots are always real

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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

OpenStudy (anonymous):

oh what is it then?

OpenStudy (turingtest):

I just wrote it above

OpenStudy (turingtest):

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

OpenStudy (turingtest):

so the discriminant is always \(\ge0\)

OpenStudy (turingtest):

which means the roots are always real

OpenStudy (anonymous):

that was fun. thanks

OpenStudy (turingtest):

yeah it was welcome :)

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