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November 29, 2015
Posted by **margiee (: !** on Friday, October 24, 2008 at 11:50am.

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

[please help me by starting off the problems, or explaining for each.]

- math [ i need help with these.] -
**Quidditch**, Friday, October 24, 2008 at 12:32pm2. 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.] -
**margiee (: !**, Friday, October 24, 2008 at 12:34pmx=-3 ?

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- math [ i need help with these.] -
**Quidditch**, Friday, October 24, 2008 at 12:34pmSorry, I meant to say factor not term.

If any "factor" in a product is 0 the whole product is 0.

There are 2 "factors".

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**margiee (: !**, Friday, October 24, 2008 at 12:39pmi still don't understand , i'm sorry .

- math [ i need help with these.] -
**Quidditch**, Friday, October 24, 2008 at 12:52pm2x(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.] -
**Quidditch**, Friday, October 24, 2008 at 12:55pmThe line

-0(3)

should have been

=0(3)

See below:

For x=0

2x(x+3)

=2(0)(0+3)

=0(3)

=0

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- math [ i need help with these.] -

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**drwls**, Friday, October 24, 2008 at 12:35pmAren'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.] -
**margiee (: !**, Friday, October 24, 2008 at 12:58pmmkay , thanks (:

so , what about number 3 ?

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- math [ i need help with these.] -
**drwls**, Friday, October 24, 2008 at 2:33pmFactor 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.