Quick question. (2 Viewers)

[currysauce]

Show that the roots of the equation

4(m+1)x² -4(m-1)x - 3=0 (m not equal to -1) are real for all real m


SORRY

 

[Will Hunting]

Show that the roots of the equation are what?

 

[Trev]

Show that the roots of the equation.......what is the rest?

 

[Estel]

for real roots discriminant >= 0
discriminant:
16(m-1)^2+48(m+1)
= 16m^2 + 16m + 64
= 16(m+1/2)^2 + 60 >=0 for all m

 

[currysauce]

could u write out this in more detail, i don't understand ur last line of working

 

[Estel]

The last line is completing the square.
A square is always >= 0, hence a square + a positive constant is always >= 0.

 

[currysauce]

could u simplify the 16m² +16m +64 to m²+m+4

how would u do ur trick then?

 

[Trev]

In the last line of Estel's working, he just did the following:
= 16m^2 + 16m + 64
= 16(m² + m) + 64
= 16(m + 1/2)² - (16*1/4) + 64
= 16(m + 1/2)² - 4 + 60
= 16(m+1/2)^2 + 60

 

[Trev]

Just wondering Estel, what are your other subjects apart from extension 2 mathematics?

 

[Estel]

currysauce, you can certainly factorise, but m^2 + m + 4 is definitely not 16m^2 + 16m^2 + 64 "simplified"

4u english, 3u hist (mod) and well SOR doesn't really count

 

[Slidey]

The last line is completing the square.
A square is always >= 0, hence a square + a positive constant is always >= 0.

Exclusively greater than?

 

[Estel]

I wanted the final statement to be the same as the statement I set out to prove.
Me being pedantic

 

[Will Hunting]

Estel,

Consider m = 1
Then, F(x) = 4(1+1)x^2 -4(1-1)x - 3 = 0
8x^2 - 3 = 0 (a difference of 2 squares, 2root2x and root3)
Therefore, x = root3/2root2, -root3/2root2 = root6/4, -root6/4

m = 1 into your equation, however, gives

discriminant = 16(1+1/2)^2 + 60 = 96
whence x = (-0 +- 8)/16 = 1/2, -1/2 (not equal to root6/4, -root6/4)

Your statement seems to hold for all other real values of m, m not equal to +-1, i.e. when a,b,c are not equal to 0
(You can omit the = from the >= sign as F(x) always has distinct roots)

currysauce, are you sure the question didn't read m not equal to +- 1?

 

[currysauce]

just -1


thanks for the help guys as well

 

[Estel]

Estel,

Consider m = 1
Then, F(x) = 4(1+1)x^2 -4(1-1)x - 3 = 0
8x^2 - 3 = 0 (a difference of 2 squares, 2root2x and root3)
Therefore, x = root3/2root2, -root3/2root2 = root6/4, -root6/4

m = 1 into your equation, however, gives

discriminant = 16(1+1/2)^2 + 60 = 96
whence x = (-0 +- 8)/16 = 1/2, -1/2 (not equal to root6/4, -root6/4)

Your statement seems to hold for all other real values of m, m not equal to +-1, i.e. when a,b,c are not equal to 0
(You can omit the = from the >= sign as F(x) always has distinct roots)

currysauce, are you sure the question didn't read m not equal to +- 1?

Firstly, I don't have a clue what you're on about.
If you're saying what I think you are...
if m=1, x= +- rt(3/8)
and in solving for x using the discrimant you get (+-rt 96)/16; I don't know where you got 1/2 from.
The statement is in fact true for all m.

Also, I set out to prove that the roots were real, hence I set out to prove >=. a>b implies a>=b true enough, but it's best to have equivalent final statements.
To each his own.

 

[currysauce]

how about the prob qu's

 

[Will Hunting]

solving for x using the discrimant you get (+-rt 96)/16

The sol'ns 1/2, -1/2 are incorrect. The statement is, on review, true for all real m, provided a couple of amendments are made.

Other than that, there are no probs with my reasoning, so I'm not sure what's eating you.

If you're saying what I think you are...
if m=1, x= +- rt(3/8)

What I'm getting at there should be crystal clear. It's a simple application of difference of two squares to solve a quadratic for x. Or, otherwise, if its easier,

8x^2 - 3 = 0
8x^2 = 3
x^2 = 3/8
x = +- rt(3/8)

Also, I set out to prove that the roots were real, hence I set out to prove >=.

I don't understand your wording. Did you mean to type "hence I set out to prove that the discriminant is >= 0"?

On that point, whilst proving that the discriminant (A) >= 0 does, further, prove real roots, so too does simply proving that A > 0. This is because positive numbers, with the exclusion of 0, are a subset of the real field. In this light, you should have realised, prior to answering, that F(x) would never have equal roots (and, therefore, A = 0) for any real m and could have, from that, sidestepped a false response. I'll, therefore, say this for the second time - the = from the >= can be omitted.

it's best to have equivalent final statements

If you think so, then why don't you? You set out to prove real roots, as you say, but then you made the false statement, A >= 0 when all you needed to do was prove A > 0 for real roots. You should rethink both your initial, and your concluding statement, man, 'cause you're contradicting yourself.

a>b implies a>=b

No it doesn't.

Anyways, it's all good dude and it's good to critique back and forth like this. The statement you got was tops, except for just a little niggly thing. It's no biggie, but I guess it's just "me being pedantic"

 

[Trev]

Haha, math duel, who will win - I got bets on Estel.

 

[Slidey]

I'm not getting involved in the argument, but...

I think Estel is trying to argue that since it is logically P v Q, than even P and never Q, P v Q returns true. Specifically:

x >= 0 is (x>0) v (x=0). For any real value of x (even if x is never zero), this statement will return true. That is to say: For x>0, it follows that x>=0, with some redundancy. Personally, though, I'd never put that down in a test, since I'd have to haggle for ages when I got that question back with a cross on it.

Interestingly if we really wanted exclusivity it could be represented by (x > 0) ^ (x != 0), which reads as "x is greater than zero, and x is not zero".

Will's argument that > is a subset of >=, hence > does not imply >= is an interesting one and the one I'd have intuitively gone for. I'll have to think about it some more as I am not happy with the redundancy which occurs with Estel's argument.

 

[Templar]

No it doesn't.

Actually, it does.