Completing the square

**llkkjj24** · December 20th 2009, 09:26 AM

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 $1 \pm \sqrt{\frac{2}{3}}$

**Krahl** · December 20th 2009, 09:38 AM

a)

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

**Krahl** · December 20th 2009, 09:45 AM

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.

**llkkjj24** · December 21st 2009, 09:21 AM

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

**jgv115** · December 21st 2009, 10:51 PM

stuffed the question up...

**jgv115** · December 21st 2009, 11:00 PM

$2x^2 -3x -3 = 0$

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?

Thread: Completing the square

LinkBack

Thread Tools

Display

Completing the square

Similar Math Help Forum Discussions

Completing the square.

completing the square

completing the square?

Completing The Square

Completing the Square

Search Tags