polynomials question (1 Viewer)

[Hermes1]

Prove that if a not equal to c there are always two real values of k which will make:

a perfect square

 

[woga3]

come on , that's a 2unit question (maybe slightly harder , maybe Q8-Q9 worthy in a 2unit hsc ) .Get that out of 4unit lols.

Do you really think actuary is for you?

I haven't done any actual working for it , but it looks pretty easy at first glance.

I am guessing you know who this is .

Also I might have someone to tutor this saturday. Not a hsc student but a uni student, to review easy 2unit stuff. Calculus, Rates, Series, Financial Maths.

He has emailed me back 3times (which is a record!, most email me and then I never hear from then again.) . Said he was going to call me tonight.

 

[Hermes1]

come on , that's a 2unit question (maybe slightly harder , maybe Q8-Q9 worthy in a 2unit hsc ) .Get that out of 4unit lols.

Do you really think actuary is for you?

I haven't done any actual working for it , but it looks pretty easy at first glance.

I am guessing you know who this is .

Also I might have someone to tutor this saturday. Not a hsc student but a uni student, to review easy 2unit stuff. Calculus, Rates, Series, Financial Maths.

He has emailed me back 3times (which is a record!, most email me and then I never hear from then again.) . Said he was going to call me tonight.

lol just answer the question or go away.

 

[woga3]

lol just answer the question or go away.

ok, I gotta get home away, seeya

 

[nightweaver066]

May be wrong but don't you expand, group the coefficients so it looks more like a quadratic.

Then you use discriminant and for perfect square, discriminant = 0

Then whilst you make k the subject, you'll end up with a + or - square root.

 

[woga3]

May be wrong but don't you expand, group the coefficients so it looks more like a quadratic.

Then you use discriminant and for perfect square, discriminant = 0

Then whilst you make k the subject, you'll end up with a + or - square root.

someones a bit more switched on, and they are doing their hsc next year lols!

 

[Hermes1]

May be wrong but don't you expand, group the coefficients so it looks more like a quadratic.

Then you use discriminant and for perfect square, discriminant = 0

Then whilst you make k the subject, you'll end up with a + or - square root.

that's what i tried first. but ill try it again. tnx.

 

[Hermes1]

someones a bit more switched on, and they are doing their hsc next year lols!

is this goodwin?. im coming to UNSW next year mate, and im goin to get u.

 

[nightweaver066]

The answer i ended up with:

 

[Hermes1]

The answer i ended up with:

but u hav k on both LHS and RHS.

 

[nightweaver066]

but u hav k on both LHS and RHS.

Meant to be . Sorry about that.

 

[woga3]

4b^2 -4(a+k)(c+k) =0

then expand that as a quadratic in k , and find its discriminant again to find how many solns it has. The leading coeffiecent is negative so you need positive discriminant for it to have two solns for k

you can tell its positive by using the squares in it , so to speak.

There , im not such an asshole lols

 

[Hermes1]

Meant to be . Sorry about that.

tnx. for that.

 

[Hermes1]

4b^2 -4(a+k)(c+k) =0

then expand that as a quadratic in k , and find its discriminant again to find how many solns it has. The leading coeffiecent is negative so you need positive discriminant for it to have two solns for k

you can tell its positive by using the squares in it , so to speak.

There , im not such an asshole lols

yes ur not so bad after all.

 

[woga3]

LHS = 4b^2 - 4 ( ac +ak +ck +k^2) = -4k^2 -4k (a+c) +4b^2 -4ac

discriminant = 16(a+c)^2 -4(-4)(4(b^2-ac) ) = 16(a+c)^2 + 64(b^2-4ac) do some shit with that.

 

[Drongoski]

Prove that if a not equal to c there are always two real values of k which will make:

a perfect square

I saw this at 3.20 pm but could not post solution as my internet was down and I had to go out.


If has a repeated root then the quadratic in x is a perfect square.

This eqn is:

Now we have repeated root if





a quadratic eqn in k with 2 distinct roots if




but therefore we need:




i.e.

That means, if a =/= c the original quadratic is a perfect square for 2 distinct real values of k.

 

[Hermes1]

I saw this at 3.20 pm but could not post solution as my internet was down and I had to go out.


If has a repeated root then the quadratic in x is a perfect square.

This eqn is:

Now we have repeated root if





a quadratic eqn in k with 2 distinct roots if




but therefore we need:




i.e.

That means, if a =/= c the original quadratic is a perfect square for 2 distinct real values of k.

tnx so much drongoski, this was the solution i was lookin for. also is the 41 in the sixth last line supposed to be a 4?.

 

[Drongoski]

It's 4 indeed. I had intended to insert an "x" between the 4 and the 1.

I typed \timex when I intended \times.

 

[Hermes1]

It's 4 indeed. I had intended to insert an "x" between the 4 and the 1.

I typed \timex when I intended \times.

ah ok. tnx again drongoski. this question was good for a bit of revision for myself of the discriminant and its properties.

 

[nightweaver066]

Good for me learning a few things too, thanks Drongoski.