1. Oct 11, 2009

### zebra1707

1. The problem statement, all variables and given/known data

2x^2 + 7x - 5 = A(x-1)^2 + Bx + C

2. Relevant equations

N/a

3. The attempt at a solution

This is an interesting equation as regardless of the whether I substitute x = 0 or x = 1 i still encounter the problem of not be able to make the calculation any easier. I have not been able to find any examples for guidence. Can someone assist.

Cheers

Last edited: Oct 11, 2009
2. Oct 11, 2009

### Dick

That's just an equation. What's the actual problem? Are you supposed to find A, B, and C given the equation is true for all x? That's just a guess.

3. Oct 11, 2009

### zebra1707

yes, sorry - need to find A, B and C

Cheers

4. Oct 11, 2009

### Dick

Ok, then if ax^2+bx+c=0 for all x then a=0 and b=0 and c=0. Expand everything and move everything to one side. Set the coefficients of powers of x to zero and solve for A, B and C.

5. Oct 12, 2009

### HallsofIvy

Staff Emeritus
Or do exactly what you suggested in your original post.

If x= 0, this becomes -5= A+ C. If x= 1, this becomes 4= B+ C. Since there are three unknown numbers, you need a third equation: If x= -1, this becomes -10= 4A- B+ C. Three equations to solve for A, B, and C. From equation one, A= -5-C. From equation 2, B= 4- C. Replace A and B with those in the third equation and solve for C.

6. Oct 12, 2009

### zebra1707

I ended up with A=2, B=11 and C= -7

Substituting the figures in to the equation - it balances with the R.H.S.

I think that I have solved it - but happy for comment if I have it wrong.

Cheers

7. Oct 12, 2009

### Staff: Mentor

A(x-1)2 + Bx + C = Ax2 + (B-2A)x + A+C

So at least A=2 is obviously OK, then A+C=2+C=-5, so C=-7. B-2A=B-4=7, so B is OK too.

--
methods

8. Oct 12, 2009

### zebra1707

A BIG thanks to all that have contributed to my understanding of this.

Greatly appreciated.