general form to standard form

**katt** · November 26th 2009, 03:12 PM

i need to change = -3x^2+x-2 to standard form( a(x-p)^2 +q)
can anyone show me the algebraic steps through completing the square?

**Raoh** · November 26th 2009, 03:28 PM

$ax^{2}+bx+c=a\left [ \left ( x+\frac{b}{2a}\right )^{2}-\frac{\Delta}{4a^{2}}\right ]$

**katt** · November 26th 2009, 03:48 PM

i don't get your second equation already = a[(x+b/2a)^2 -delta/4a^2]

**dedust** · November 26th 2009, 05:56 PM

this is the details

$ax^{2}+bx+c=0$

$x^{2}+\frac{b}{a}x+\frac{c}{a}=0$

$\left( x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}\right) +\frac{c}{a}%<br /> -\frac{b^{2}}{4a^{2}}=0$

$\left( x+\frac{b}{2a}\right) ^{2}+\frac{c}{a}-\frac{b^{2}}{4a^{2}}=0$

$\left( x+\frac{b}{2a}\right) ^{2}+\frac{4ac-b^{2}}{4a^{2}}=0$

**Raoh** · November 27th 2009, 01:04 AM

here's what u need,

Homeomath : forme canonique de ax² + bx + c
forget about the writing

**Raoh** · November 29th 2009, 03:43 AM

okay,
$-3x^2+x-2$ = $-3\left ( x^2-\frac{1}{3}x+\frac{2}{3} \right )=-3\left ( (x-\frac{1}{6})^2 -\frac{1}{36}+\frac{2}{3}\right )$ = $-3\left ( (x-\frac{1}{6})^2+\frac{23}{36} \right )$

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