TrueTears question thread

[TrueTears]

Then finally "Hence without differentiating, show that , where and are real numbers, is a general solution of (A)."

How to go about this?

[zzdfa]

Next part of the question says:

(B) could also have 1 repeated root, say m.

Show that 2am+b=0

Do I just simply implicit differentiate both sides of equation (B) ?

You get the correct answer but I don't think that's correct method, the correct reasoning is:
let y=(B)
then find dy/dx=2am+b
For a quadratic, If m has a repeated root, you know it means the turning point must coincide with the intersection of the x-axis.
Therefore solving
0=2am+b will give you the x-coord of the maximum/minimum which is also the root.

OR you could use the fact that the 2 roots of a quadratic equation are:

And for them to be equal, the stuff under the square root has to equal 0. so you have
m=-b/2a, rearrange and youre done.

Then finally "Hence without differentiating, show that , where and are real numbers, is a general solution of (A)."

How to go about this?

Well we know that when you sub the bit into [A] the equation equals to 0, so we can ignore that, since differentiation/integration is linear. and now im stuck. are you sure you gave us all the information?

coz atm this is what it's like:

 

   (where m=-b/2a)


and we're not allowed to differentiate.

[TrueTears]

Thanks for your help.

Yeah that is all the information.

btw is my method for first part correct? I'm just wondering if there's any other method.

[zzdfa]

yea that's the way i'd do it, basically using the fact that 0+0=0

[TrueTears]

Yeah thanks.

Actually I think the question says given that is a solution of (A) [Part of the Q was cut off from the page]

Then show that is a general solution of (A).

How would you approach it now?

[zzdfa]

lol, then it's just the same as the first question, key thing is
http://en.wikipedia.org/wiki/Linearity_of_differentiation    (try to ignore all the scary greek letters, you probably know what it is already)

If is a solution then is a solution since . Then  is also a solution since . you can fill in the details.

[NE2000]

Yeah thanks.

Actually I think the question says given that is a solution of (A) [Part of the Q was cut off from the page]

Then show that is a general solution of (A).

How would you approach it now?

Won't you do it the same way as the first part or am I missing something :S

EDIT: zzdfa comes in before me

[TrueTears]

Oh yeah thanks, so if you expand the double prime " into the brackets it would be ?

[TrueTears]

Ah nvm I got it. Thanks all.

[TrueTears]

What about if (B) had no real roots, and since the coefficients are real then 2 complex roots would be [complex conjugate]. Using the fact that v + wi is a root of (B) show that

I just subbed m = v+wi into (B) but it doesn't work.

[zzdfa]

sub in m=v+wi, and after some algebra you get


thus
to show that 
all we need to do is to  show that
look familiar? remember the real part of v+wi comes from the stuff outside the square root.

[TrueTears]

True, 2av+b = 0 , how to show that?

[zzdfa]

the 2 roots of a quadratic equation are:



in both cases,

[TrueTears]

Oh I see, then what is the imaginary part 'w' of the quadratic equation?

Is it the square root itself or under the square root.

[zzdfa]

i'm not sure what you're asking:
solution to a quadratic is:

set

and if the stuff under sqrt is negative, you get an imaginary component with: