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15 votes
15 votes
Given that 2x^2-5 = a(x+2)^2 _ b(x+1)+x for all values of X, find the value of a, of b and of c

User Guildencrantz
by
2.8k points

1 Answer

25 votes
25 votes

It looks like you're asking to solve for the coefficients
a,b,c such that


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

Expanding the right side gives


2x^2 - 5 = ax^2 + 4ax + 4a - bx - b + c


2x^2 - 5 = ax^2 + (4a - b)x + (4a - b + c)

The coefficients on both sides must be equal, so


\begin{cases}a = 2 \\ 4a-b = 0 \\ 4a - b + c = -5\end{cases}

Solving the system yields
\boxed{a=2 \implies b = 8 \implies c = -5}.

User Kklo
by
3.2k points
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