completing the square

**offahengaway and chips** · September 22nd 2008, 06:36 AM

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?

**kalagota** · September 22nd 2008, 07:21 AM

Originally Posted by offahengaway and chips

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?

$5(x^2+4x + r) - 12 -5r$

can you work from here? find that $r$ so that the thing in the parenthesis is a perfect square..

**offahengaway and chips** · September 22nd 2008, 07:26 AM

Still a bit confused here, is r=c?

**Shyam** · September 22nd 2008, 07:39 AM

Originally Posted by offahengaway and chips

Still a bit confused here, is r=c?

$<br /> =5x^2+20x-12$

$=5(x^ 2+4x)-12$

$=5(x^2 + 4x +4 -4) -12$

$=5(x+2)^2-20-12$

Now, finish up.

**offahengaway and chips** · September 22nd 2008, 07:47 AM

Just to double check is c= -32?

**Shyam** · September 22nd 2008, 08:33 AM

yes, it is -32.

**bkarpuz** · September 22nd 2008, 08:58 AM

In general, for $A,B,C\in\mathbb{R}$ with $A\neq0$ , we have
$Ax^{2}+Bx+C=A\Bigg(x^{2}+\frac{B}{A}x\Bigg)+C$
...................._ $=A\Bigg(x^{2}+\frac{B}{A}x+\bigg(\frac{B}{2A}\bigg )^{2}-\bigg(\frac{B}{2A}\bigg)^{2}\Bigg)+C$
...................._ $=A\Bigg(x^{2}+\frac{B}{A}x+\bigg(\frac{B}{2A}\bigg )^{2}\Bigg)+C-\frac{1}{A}\bigg(\frac{B}{2}\bigg)^{2}$
...................._ $=A\Bigg(x+\frac{B}{2A}\Bigg)^{2}+C-\frac{1}{A}\bigg(\frac{B}{2}\bigg)^{2}$

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