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

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Homework Help Overview

The discussion revolves around solving the quadratic identity represented by the equation 2x^2 + 7x - 5 = A(x-1)^2 + Bx + C. Participants are tasked with determining the values of A, B, and C such that the equation holds true for all values of x.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore different methods to find A, B, and C, including substituting specific values for x and expanding the equation. Some question the clarity of the problem statement and seek confirmation on the goal of finding coefficients.

Discussion Status

There is an ongoing exploration of various approaches to solve for A, B, and C. Some participants have provided specific equations derived from substitutions, while others have confirmed certain values. However, there is no explicit consensus on the final values yet, and further verification is encouraged.

Contextual Notes

Participants note the need for a third equation to solve for the three unknowns, indicating that the problem requires careful consideration of the relationships between the coefficients.

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

Cheers


Dick said:
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.
 
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.
 
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.
 
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
 
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
 

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