
Completing the square
solve the following quadratic equation , solution in the form $\displaystyle a \pm b \sqrt{n}$
a) $\displaystyle 2x^2 3x 3 = 0 $
b) $\displaystyle 3x^2 6x +1 = 0$
thanks
ps : for a i got completely messed up.
for b i got $\displaystyle 1 \pm \frac{\sqrt{2}}{3}$ but the answer from the maths book was $\displaystyle 1 \pm \sqrt{\frac{2}{3}}$

a)
$\displaystyle 2x^2 3x 3 = 0$
$\displaystyle 2(x^2 \frac{3}{2}x) 3= 0$
$\displaystyle 2((x\frac{3}{4})^2(\frac{3}{4})^2) 3= 0$
$\displaystyle 2((x\frac{3}{4})^2)2(\frac{3}{4})^2 3= 0$

This is the procedure;
$\displaystyle ax^2+bx+c=0$
$\displaystyle a(x^2+\frac{b}{a}x)+c=0$
$\displaystyle a((x+\frac{b}{2a})^2(\frac{b}{2a})^2)+c=0$
$\displaystyle a((x+\frac{b}{2a})^2)a(\frac{b}{2a})^2+c=0$
$\displaystyle a((x+\frac{b}{2a})^2)=a(\frac{b}{2a})^2c$
$\displaystyle (x+\frac{b}{2a})^2=(\frac{b}{2a})^2\frac{c}{a}$
and solving this equation for x you get the quadratic formula.

thanks for the help, but i found out that is more easy to just take about the coeffecient of x out as a common factor, complete the sqaure, then just multiply it back in

stuffed the question up...

$\displaystyle 2x^2 3x 3 = 0 $
Same thing. Take out 2 as a common factor then divide coefficient by 2 and square.
$\displaystyle x^2\frac{3}{2}x  \frac{3}{2} = 0 $
Coefficient divide by 2:
$\displaystyle \frac{3}{2} * \frac{1}{2} = \frac{3}{4} $
Square:
$\displaystyle (\frac{3}{4})^2 = \frac{9}{16} $
Now complete the square by dividing the coefficient by 2
$\displaystyle (x\frac{3}{4})^2  \frac{33}{16} =0 $
Can you do it from here.
Do you understand everything I'm doing?