Completing the square

• December 20th 2009, 09:26 AM
llkkjj24
Completing the square
solve the following quadratic equation , solution in the form $a \pm b \sqrt{n}$

a) $2x^2 -3x -3 = 0$
b) $3x^2 -6x +1 = 0$

thanks

ps : for a i got completely messed up.
for b i got $1 \pm \frac{\sqrt{2}}{3}$ but the answer from the maths book was $1 \pm \sqrt{\frac{2}{3}}$
• December 20th 2009, 09:38 AM
Krahl
a)

$2x^2 -3x -3 = 0$
$2(x^2 -\frac{3}{2}x) -3= 0$
$2((x-\frac{3}{4})^2-(\frac{3}{4})^2) -3= 0$
$2((x-\frac{3}{4})^2)-2(\frac{3}{4})^2 -3= 0$
• December 20th 2009, 09:45 AM
Krahl
This is the procedure;

$ax^2+bx+c=0$
$a(x^2+\frac{b}{a}x)+c=0$
$a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2)+c=0$
$a((x+\frac{b}{2a})^2)-a(\frac{b}{2a})^2+c=0$
$a((x+\frac{b}{2a})^2)=a(\frac{b}{2a})^2-c$
$(x+\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}$
and solving this equation for x you get the quadratic formula.
• December 21st 2009, 09:21 AM
llkkjj24
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
• December 21st 2009, 10:51 PM
jgv115
stuffed the question up...
• December 21st 2009, 11:00 PM
jgv115
$2x^2 -3x -3 = 0$

Same thing. Take out 2 as a common factor then divide coefficient by 2 and square.

$x^2-\frac{3}{2}x - \frac{3}{2} = 0$

Coefficient divide by 2:

$\frac{3}{2} * \frac{1}{2} = \frac{3}{4}$

Square:

$(\frac{3}{4})^2 = \frac{9}{16}$

Now complete the square by dividing the coefficient by 2

$(x-\frac{3}{4})^2 - \frac{33}{16} =0$

Can you do it from here.

Do you understand everything I'm doing?