• Apr 13th 2008, 10:47 AM
gumi
solve the following quadratics by factoring

x^2=6x

x^2=6x-8

6x^2-13x+6=0

5x^2-17x+14=0

12x^2+x=1
• Apr 13th 2008, 10:49 AM
Mathstud28
Ok
Quote:

Originally Posted by gumi
solve the following quadratics by factoring

x^2=6x

x^2=6x-8

6x^2-13x+6=0

5x^2-17x+14=0

12x^2+x=1

Why dont you show us some work and we will critique you...for the first one I will give you that $(x-3)^2=9$
• Apr 13th 2008, 11:05 AM
gumi
for the 2nd one i got
(x-3)^2=1

is that right?
• Apr 13th 2008, 11:07 AM
Mathstud28
Haha
Quote:

Originally Posted by gumi
for the 2nd one i got
(x-3)^2=1

is that right?

Yes that is correct...very good...but it can be simpler factor it into $(x-4)(x-2)$
• Apr 13th 2008, 11:08 AM
Moo
Hello,

For the first one, do you really need to complete the square ? oO It's only asked to factorise ^^

x²=6x
x²-6x=0
x(x-6)=0

So...
• Apr 13th 2008, 11:09 AM
Mathstud28
yeah I know
Quote:

Originally Posted by Moo
Hello,

For the first one, do you really need to complete the square ? oO It's only asked to factorise ^^

x²=6x
x²-6x=0
x(x-6)=0

So...

But on another post of theirs it asked for completing the square so I just took the liberty of assuming that is what she meant
• Apr 13th 2008, 11:25 AM
gumi
the 3rd one i got
(2 + -3x)(3 + -2x) = 0

is that right?