# integrated algebra

posted by on .

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 - ,

x (x-3) = 0
x = 0 or x = 3

• integrated algebra - ,

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 - ,

No, because -0x-0x is not -3x
You missed the x = 3 root entirely.
Please do it my way :)

• integrated algebra - ,

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 - ,

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 - ,

do i always have to mention the ors

• integrated algebra - ,

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 - ,

oh, okay, thank you!

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