1. ## College Algebra

Need Help getting started on the following problems, it's been almost 10 years since I have had any algebra, but it was always my favorite subject, so if I know I'm doing it correctly, I will catch back on.

1) solve for a: a^2 + 4a=45

2) Let f(x)= x^2+2x-2 and g(x)= 2-3x. Find the following:
a) f(-1)
b) g(-3)
c) f(a+b)

3)solve for x: (2/x-1) +1=(2/x^2-x)

4) sove this system of equations:

5x+4y=-8
2x-y=-11

If anyone can help, it would be greatly appreciated!

2. ## Re: College Algebra

$a^2+4a \ = \ 45, \ \implies \ a^2+4a-45 \ = \ 0.$

$Hence, \ using \ grouping, \ we \ get \ a^2+9a-5a-45 \ = \ 0$

$Then \ factoring, \ we \ get \ a(a+9)-5(a+9), \ gives \ (a+9)(a-5) \ = \ 0$

$Therefore, \ by \ the \ zero \ property \ rule \ a \ = \ -9 \ or \ a \ = \ 5.$

$I&#39;ll \ leave \ the \ check \ up \ to \ you.$

3. ## Re: College Algebra

I get all of this except after you factor, you said a= -9 or 5 and when I factored I used 9, -5 thus this would = the positive 4, can you explain, thanks.

4. Originally Posted by twhitley

I get all of this except after you factor, you said a= -9 or 5 ? Glenn is talking about the solutions, here, not the factorization

and when I factored I used 9, -5 ? This is correct; Glenn also used 9 and -5 in the factorization

thus this would = the positive 4 Your use of the pronoun "this" is ambiguous.

You're not adding the two numbers in the factorization together, are you?

5. Originally Posted by twhitley

1) solve for a: a^2 + 4a=45
Do you understand the meaning of the phrase "solve for a" ?

If not, then please tell us right away because that is where we would need to begin helping you.

6. ## Re: College Algebra

no I'm not adding them , I remember learning to factor and if it added up to the 2nd number then it was right, with that disregarded.
Solving for a, I think of it as solving for x or y, is this correct?
there is an (a) and (a^2) so I guess this is where I'm confused.
Thanks.

7. ## Re: College Algebra

$(a+9)(a-5) \ = \ 0$

$Now, \ if \ a \ = \ -9 \ or \ if \ a \ = \ 5, \ the \ above \ equation \ rings \ true.$

$Check: \ (-9+9)(-9-5) \ = \ 0, \ 0(-14) \ = \ 0, \ 0 \ = \ 0,$

$and \ (5+9)(5-5) \ = \ 0, \ (14)(0) \ = \ 0, \ 0 \ = \ 0$

8. ## Re: College Algebra

Originally Posted by twhitley
no I'm not adding them , I remember learning to factor and if it added up to the 2nd number then it was right, with that disregarded.
Solving for a, I think of it as solving for x or y, is this correct?
there is an (a) and (a^2) so I guess this is where I'm confused.
Thanks.
If the product of two or more factors is 0, then at least one of those factors must be 0.

Symbolically: if a*b = 0, then a = 0 or b = 0.

We often apply this principal to solve quadratic (or higher degree) equations which can be factored. For example, if we have to solve

x[sup:tthh96kl]3[/sup:tthh96kl] - 5x[sup:tthh96kl]2[/sup:tthh96kl] + 6x = 0, we might start by factoring:

x(x[sup:tthh96kl]2[/sup:tthh96kl] - 5x + 6) = 0
x(x - 2)(x - 3) = 0

Now, since we have three factors whose product is 0, we know that at least one of those factors must have a value of 0.

So....since x(x - 2)(x - 3) = 0, we can say
x = 0
OR
x - 2 = 0, which means x = 2
OR
x - 3 = 0, which means x = 3.

So we have three solutions: x = 0 or x = 2 or x = 3

9. ## Re: College Algebra

ok, I get that , does this mean to answer the question I would put both down, a= -9 or a=5, my thought process was to only have one answer for a.
thanks!

10. ## Re: College Algebra

cool , i got it , thank all of you soooo much, have you looked at any of the others?

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