complete squares

**Alienis Back** · October 27th 2008, 05:31 AM

Hi there. I need to know where is my mistake so I can correct it. Hope you can help me.

**flyingsquirrel** · October 27th 2008, 05:41 AM

Hi

The problem is that $a^2+5a+1^2$ is not a perfect square. A perfect square is something like $x^2+2xy+y^2=(x+y)^2$ . If you choose $x=a$ and $y=1$ , the term $2xy$ equals $2a$ , not $5a$ ... it doesn't work.

**earboth** · October 27th 2008, 05:46 AM

Originally Posted by Alienis Back

Hi there. I need to know where is my mistake so I can correct it. Hope you can help me.

Unfortunately you made a mistake when factorizing

$a^2+5a+1 = \underbrace{a^2+5a+{\bold{\color{red}\dfrac{25}4 - \dfrac{25}4}} + 1}_{completing\ the\ square} = \underbrace{\left(a+\dfrac52\right)^2-\dfrac{21}4}_{sum\ of\ squares} =$ $\underbrace{\left[\left(a+\dfrac52\right)+\sqrt{\dfrac{21}4}\right] \left[\left(a+\dfrac52\right)-\sqrt{\dfrac{21}4}\right]}_{factorized}$

And now go ahead!

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