# math [ i need help with these.]

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2. 2x(x+3)=0
2x^2+3x=0
[i don't know what to do.]

3. y^2+8y=0
4. x^2+3x=28
5.2x^2+9x-5=0
2x^2+9x=-5

• math [ i need help with these.] -

2. 2x(x+3)=0
There are 2 terms: 2x and (x+3).
If any term in a product is 0 the whole product is 0. So, what value of x would make 2x=0 and what value of x would make x+3 = 0?

• math [ i need help with these.] -

x=-3 ?

• math [ i need help with these.] -

Sorry, I meant to say factor not term.
If any "factor" in a product is 0 the whole product is 0.
There are 2 "factors".

• math [ i need help with these.] -

i still don't understand , i'm sorry .

• math [ i need help with these.] -

2x(x+3)=0
I am assuming that you need to find the values of x that will be true for the above.
When you multiply several items (each of these is called a factor) together the answer (called the product) is 0 if ANY of them are 0.
Example:
multiply (a)(b)(c)
If any of them, a or b or c is 0, then multiplying them always gives 0.
The problem has 2 factors:
2x
(x+3)
If either of these is 0 then 2x(x+3) is 0.
The answers are 0 and -3.
It is always a good idea to substitute back to check.
For x=0
2x(x+3)
=2(0)(0+3)
-0(3)
=0

For x=-3
2(-3)(-3 + 3)
=(-6)(0)
=0

• math [ i need help with these.] -

The line
-0(3)
should have been
=0(3)
See below:
For x=0
2x(x+3)
=2(0)(0+3)
=0(3)
=0

• math [ i need help with these.] -

Aren't you supposed to solve these equations for x?

For #2, if
2x(x+3)=0, then either x = 0 or x = -3 will satisfy equation. Those are the solutions,

For #4, note that x^2 + 3x - 28 =
(x+7)(x-4). Since this is zero, then x = -7 and x = 4 satisfy the original equation, and are solutions

The two equations that you wrote for #5 are inconsistent. If the first one is correct, then what you need to do is factor 2x^2+9x-5
Try (2x-1)(x+5)
If that equals zero, then either factor is zero, and that tells you what x is.

• math [ i need help with these.] -

mkay , thanks (:

so , what about number 3 ?

• math [ i need help with these.] -

Factor y^2 + 8y = 0 into
y(y+8) = 0, and get the solutions from the factors. We are trying to teach you how to do these by yourself.

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