# general form to standard form

• Nov 26th 2009, 03:12 PM
katt
general form to standard form
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?
• Nov 26th 2009, 03:28 PM
Raoh
$\displaystyle ax^{2}+bx+c=a\left [ \left ( x+\frac{b}{2a}\right )^{2}-\frac{\Delta}{4a^{2}}\right ]$
• Nov 26th 2009, 03:48 PM
katt
i don't get your second equation already = a[(x+b/2a)^2 -delta/4a^2] (Shake)
• Nov 26th 2009, 05:56 PM
dedust
this is the details

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

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

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

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

$\displaystyle \left( x+\frac{b}{2a}\right) ^{2}+\frac{4ac-b^{2}}{4a^{2}}=0$
• Nov 27th 2009, 01:04 AM
Raoh
• Nov 29th 2009, 03:43 AM
Raoh
okay,
$\displaystyle -3x^2+x-2$ = $\displaystyle -3\left ( x^2-\frac{1}{3}x+\frac{2}{3} \right )=-3\left ( (x-\frac{1}{6})^2 -\frac{1}{36}+\frac{2}{3}\right )$=$\displaystyle -3\left ( (x-\frac{1}{6})^2+\frac{23}{36} \right )$
(Happy)