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Please help and explain :)

Find constants a, b an c such that, for all values of x.

3x^2-5x+1 = a(x+b)^2+c
Original post by leigh.xoxo
Find constants a, b an c such that, for all values of x.

3x^2-5x+1 = a(x+b)^2+c


what have you done so far?
Reply 2
Original post by TenOfThem
what have you done so far?


don't think this is right but all i've done so far is
x(3x+5)^2-25+1
(edited 9 years ago)
Original post by leigh.xoxo
don't think this is right but all i've done so far is
x(3x+5)^2-2+1


Do you know how to complete the square

OR

Have you been taught how to compare coefficients
Reply 4
Original post by TenOfThem
Do you know how to complete the square

OR

Have you been taught how to compare coefficients


We were taught how to complete the square, but not been shown how to do it with a number in front of the x.
Original post by leigh.xoxo
We were taught how to complete the square, but not been shown how to do it with a number in front of the x.


You start with 3[x253x+13]3[x^2 - \dfrac{5}{3}x + \dfrac{1}{3}]

Can you complete the square on the expression in the bracket?
Reply 6
Original post by TenOfThem
Do you know how to complete the square

OR

Have you been taught how to compare coefficients


yeah im clueless but gave it ago
x(x-5/6)^2- 25/36 + 1/3
Original post by TenOfThem
Do you know how to complete the square

OR

Have you been taught how to compare coefficients

How could you do this by comparing coefficients?
I've heard of this when someone talks of the factor theorem. What is it?
Original post by leigh.xoxo
yeah im clueless but gave it ago
x(x-5/6)^2- 25/36 + 1/3


you have an x outside the bracket instead of the 3

3x25x+1=3[x253x+13]=3[(x56)22536+13]3x^2 - 5x + 1 = 3[x^2 -\dfrac{5}{3}x + \dfrac{1}{3}] = 3[(x -\dfrac{5}{6})^2 - \dfrac{25}{36} + \dfrac{1}{3}]


So you are getting there

Now, add the fractions and multiply by the 3
Original post by MathMeister
How could you do this by comparing coefficients?
I've heard of this when someone talks of the factor theorem. What is it?


expand the RHS and then put the coefficients for each element equal
Original post by TenOfThem
expand the RHS and then put the coefficients for each element equal

Thanks!
Reply 11
Original post by TenOfThem
you have an x outside the bracket instead of the 3

3x25x+1=3[x253x+13]=3[(x56)22536+13]3x^2 - 5x + 1 = 3[x^2 -\dfrac{5}{3}x + \dfrac{1}{3}] = 3[(x -\dfrac{5}{6})^2 - \dfrac{25}{36} + \dfrac{1}{3}]


So you are getting there

Now, add the fractions and multiply by the 3


thanks its making sense now x :smile:
Original post by leigh.xoxo
thanks its making sense now x :smile:


ok

I am now going to show you equating the coefficients as you may find it even easier

3x25x+1=a(x+b)2+c3x^2 - 5x + 1 = a(x+b)^2 + c

expanding the right hand side

3x25x+1=a(x2+2bx+b2)+c3x^2 - 5x + 1 = a(x^2 + 2bx + b^2) + c

so

3x25x+1=ax2+2abx+(ab2+c)3x^2 - 5x + 1 = ax^2 + 2abx + (ab^2+c)

Looking at the x2x^2 term you can see that 3 = a

Looking at the xx term you can se that -5 = 2ab = 6b so you can find b

Then looking at the constant you can see 1 = ab2 + c so you can find c


This is just an alternative method - either works

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