solve quadratics by factoring

**gumi** · April 13th 2008, 10:47 AM

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

**Mathstud28** · April 13th 2008, 10:49 AM

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 $(x-3)^2=9$

**gumi** · April 13th 2008, 11:05 AM

for the 2nd one i got
(x-3)^2=1

is that right?

**Mathstud28** · April 13th 2008, 11:07 AM

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 $(x-4)(x-2)$

**Moo** · April 13th 2008, 11:08 AM

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...

**Mathstud28** · April 13th 2008, 11:09 AM

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

**gumi** · April 13th 2008, 11:25 AM

the 3rd one i got
(2 + -3x)(3 + -2x) = 0

is that right?

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