Solving Quadratic Identities: 2x^2 + 7x - 5 = A(x-1)^2 + Bx + C

[zebra1707]

Homework Statement

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

Homework Equations

N/a

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

 

[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.

 

[zebra1707]

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

Cheers

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.

 

[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.

 

[HallsofIvy]

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.

 

[zebra1707]

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

Substituting the figures into 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

 

[Borek]

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.

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[zebra1707]

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

Greatly appreciated.

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.

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