Help with quadratic identities. (1 Viewer)

[Omnipotence]

Find the values of a, b and c for which m² Ξ a(m – 1)² + b(m – 2)² + c(m – 3)².

 

[Puttah]

There are various methods to do this, but starting off you should learn to equate co-efficients.

It goes something like this:

If,

then a=A, b=B, c=C

So what you need to do is expand the RHS of the equation, collect all co-efficients for each variable, , x and the constants; then equate them with the value of each co-efficient and constant on the LHS.

 

[jet]

Find the values of a, b and c for which m² Ξ a(m – 1)² + b(m – 2)² + c(m – 3)².

If you expand this, then you get

m² ≡ a[m² - 2m + 1] + b[m² - 4m + 4] + c[m² - 6m + 9]

Rearranging, you get

m² ≡ [a + b + c]m² - [2a + 4b + 6c]m + a + 4b + 9c

Thus, using the identity the previous poster gave:

a + b + c = 1
therefore a = 1 - b - c

2a + 4b + 6c = 0

2 - 2b - 2c + 4b + 6b = 0
2b + 4c = -2
b + 2c = -1 ............. (1)

a + 4b + 9c = 0
1 - b - c + 4b + 9c = 0
3b + 8c = 1
b + (8/3)c = 1/3 ..........(2)

Subtracting (1) from (2):

[(8/3) - 2]c = (1/3) + 1
(2/3)c = 4/3
c = 4/2 = 2

Substituting into (1), b = -1 - 2(2)
= -5

and a = 1 - (-5) - (2)
= 4

 

[Omnipotence]

Thanks.

 

[muzeikchun852]

If you expand this, then you get

m² ≡ a[m² - 2m + 1] + b[m² - 4m + 4] + c[m² - 6m + 9]

Rearranging, you get

m² ≡ [a + b + c]m² - [2a + 4b + 6c]m + a + 4b + 9c

Thus, using the identity the previous poster gave:

a + b + c = 1
therefore a = 1 - b - c

2a + 4b + 6c = 0

2 - 2b - 2c + 4b + 6b = 0
2b + 4c = -2
b + 2c = -1 ............. (1)

a + 4b + 9c = 0
1 - b - c + 4b + 9c = 0
3b + 8c = 1
b + (8/3)c = 1/3 ..........(2)

Subtracting (1) from (2):

[(8/3) - 2]c = (1/3) + 1
(2/3)c = 4/3
c = 4/2 = 2

Substituting into (1), b = -1 - 2(2)
= -5

and a = 1 - (-5) - (2)
= 4

wat the f is this?!
im in the 5.3..
but i didnt learn these complex thing.

 

[tommykins]

u learn it anyways

 

[study-freak]

Find the values of a, b and c for which m² Ξ a(m – 1)² + b(m – 2)² + c(m – 3)².

sub m=1
1=b+4c...(1)
sub m=2
4=a+c...(2)
sub m=3
9=4a+b...(3)
sub m=0
0=a+4b+9c...(4)

(4)-(2) gives
-4=4b+8c
-1=b+2c...(5)

(1)-(5) gives
2=2c
c=1...(6)

sub (6) into (2)
a=3

sub (6) into (1)
1=b+4
b=-3

Thus, a=3, b=-3, c=1

 

[tommykins]

nice solution above.

i would've expanded lol

 

[study-freak]

nice solution above.

i would've expanded lol

lol thanks

 

[Omnipotence]

If you expand this, then you get

m² ≡ a[m² - 2m + 1] + b[m² - 4m + 4] + c[m² - 6m + 9]

Rearranging, you get

m² ≡ [a + b + c]m² - [2a + 4b + 6c]m + a + 4b + 9c

Thus, using the identity the previous poster gave:

a + b + c = 1
therefore a = 1 - b - c

2a + 4b + 6c = 0

2 - 2b - 2c + 4b + 6b = 0
2b + 4c = -2
b + 2c = -1 ............. (1)

a + 4b + 9c = 0
1 - b - c + 4b + 9c = 0
3b + 8c = 1
b + (8/3)c = 1/3 ..........(2)

Subtracting (1) from (2):

[(8/3) - 2]c = (1/3) + 1
(2/3)c = 4/3
c = 4/2 = 2

Substituting into (1), b = -1 - 2(2)
= -5

and a = 1 - (-5) - (2)
= 4

Slight mistake, but I resolved it.

a + 4b + 9c = 0
1 - b - c + 4b + 9c = 0
3b + 8c = 1
b + (8/3)c = 1/3 ..........(2)

=> a + 4b + 9c = 0
1 - b -c + 4b+ 9c
3b + 8c = -1
b + (8/3)c = - 1/3

Yeah I expanded RHS then found corressponding co-efficent. Answer c = 1, b = - 3 and a = 3.
Now to use the quadratic function F(x) = ax^2 + bx + c to prove that a pont lies on a curve and also to prove
[m^3 over (m - n) (m - r) (m - p) + n^3 over (n - m) (n - r) (n - p)] + [r^3 over (r - m) (r - n) (r - p) + p^3 over (p - m) (p - n) (p - r)]
Do I multiply both sides, then expand? To what co-efficient though, question doesn't mention the LHS.

Anyways thank you

 

[jet]

I thought i made a mistake - I'm sick, so I'm not thinking straight lol. anyways, no worries.

 

[lpodnano]

wat the f is this?!
im in the 5.3..
but i didnt learn these complex thing.

I'm in 5.3 too. I learnt this at tutor though.

 

[ZombieApocalypse]

what? you guys are doing this in year ten?????

I literally learnt this like a week ago

whaaaaat?

 

[The Derivative]

what? you guys are doing this in year ten?????

I literally learnt this like a week ago

whaaaaat?

lol, I know right?
I was in 5.3 too

I don't think we went through the quadratic function, but we did polynomials which is super duper fun!!!!!11
I <3 Polynomial division forlyf

 

[ohexploitable]

fuck off annagurl

 

[The Derivative]

there's no need to be mean, ohexploitableeeeeeeeeeeeee/mudkip94