im having trouble working out c in this formula

The method of completing the square can be used to write the expression:

5x^2+20x-12 in the form a(x+b)^2+c where a, b and c are constants. What is the value of c?

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- Sep 22nd 2008, 06:36 AMoffahengaway and chipscompleting the square
im having trouble working out c in this formula

The method of completing the square can be used to write the expression:

5x^2+20x-12 in the form a(x+b)^2+c where a, b and c are constants. What is the value of c? - Sep 22nd 2008, 07:21 AMkalagota
- Sep 22nd 2008, 07:26 AMoffahengaway and chips
Still a bit confused here, is r=c?

- Sep 22nd 2008, 07:39 AMShyam
- Sep 22nd 2008, 07:47 AMoffahengaway and chips
Just to double check is c= -32?

- Sep 22nd 2008, 08:33 AMShyam
yes, it is -32.

- Sep 22nd 2008, 08:58 AMbkarpuz
In general, for $\displaystyle A,B,C\in\mathbb{R}$ with $\displaystyle A\neq0$, we have

$\displaystyle Ax^{2}+Bx+C=A\Bigg(x^{2}+\frac{B}{A}x\Bigg)+C$

...................._$\displaystyle =A\Bigg(x^{2}+\frac{B}{A}x+\bigg(\frac{B}{2A}\bigg )^{2}-\bigg(\frac{B}{2A}\bigg)^{2}\Bigg)+C$

...................._$\displaystyle =A\Bigg(x^{2}+\frac{B}{A}x+\bigg(\frac{B}{2A}\bigg )^{2}\Bigg)+C-\frac{1}{A}\bigg(\frac{B}{2}\bigg)^{2}$

...................._$\displaystyle =A\Bigg(x+\frac{B}{2A}\Bigg)^{2}+C-\frac{1}{A}\bigg(\frac{B}{2}\bigg)^{2}$