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^2-7x+10=0

(x-5)(x+2)= 0

x-5=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 (x-3) = 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^2-3x=0

x^2-3x+0=0

(x-0)(x+0)=0

0^2-3(0)=0

was my way right

- integrated algebra -
**Damon**, Friday, January 13, 2012 at 7:10pm
No, because -0x-0x 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

(x-3/2)^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^2-7x+10=0

(x-5)(x+2)= 0

x-5=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

(x-2)(x-2) = 0

x = 2

or

x = 2

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

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