Posted by sally on Friday, January 13, 2012 at 6:29pm.
how can you solve x^2  3x = 0 i was using a formula as:
x^27x+10=0
(x5)(x+2)= 0
x5=0
x+5 +5 x=5
I was following this formula but got stumped on this question please help.

integrated algebra  Damon, Friday, January 13, 2012 at 6:31pm
x (x3) = 0
x = 0 or x = 3

integrated algebra  sally, Friday, January 13, 2012 at 6:57pm
before looking at your answer i found the answer to be 0 too but like this:
x^23x=0
x^23x+0=0
(x0)(x+0)=0
0^23(0)=0
was my way right

integrated algebra  Damon, Friday, January 13, 2012 at 7:10pm
No, because 0x0x is not 3x
You missed the x = 3 root entirely.
Please do it my way :)

integrated algebra  Damon, Friday, January 13, 2012 at 7:15pm
If you insist on completing the square:
x^2  3x = 0
(3/2)^2 = 9/4
so
x^2  3 x + 9/4 = 9/4
(x3/2)^2 = 9/4 =3/2
x 3/2 = +/ 3/2
x = 3/2 + 3/2 = 3
pr
x = 3/2  3/2 = 0

integrated algebra  Damon, Friday, January 13, 2012 at 7:18pm
by the way
x^27x+10=0
(x5)(x+2)= 0
x5=0
x = 5
OR
x+2 = 0
x = 2

integrated algebra  sally, Friday, January 13, 2012 at 7:25pm
do i always have to mention the ors

integrated algebra  Damon, Friday, January 13, 2012 at 7:36pm
If it is a quadratic, in general it has two solutions, so you need to give both.
Sometimes, if the vertex of the parabola is exactly on the x axis, the two roots will be the same. For example:
x^2 4x + 4 = 0
(x2)(x2) = 0
x = 2
or
x = 2

integrated algebra  sally, Friday, January 13, 2012 at 7:54pm
oh, okay, thank you!
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