College Algebra

twhitley

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Joined
Aug 31, 2010
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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!
 
\(\displaystyle a^2+4a \ = \ 45, \ \implies \ a^2+4a-45 \ = \ 0.\)

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

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

\(\displaystyle Therefore, \ by \ the \ zero \ property \ rule \ a \ = \ -9 \ or \ a \ = \ 5.\)

\(\displaystyle I'll \ leave \ the \ check \ up \ to \ you.\)
 
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.
 
twhitley said:
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? :?
 
twhitley said:
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. 8-)
 
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.
 
\(\displaystyle (a+9)(a-5) \ = \ 0\)

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

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

\(\displaystyle and \ (5+9)(5-5) \ = \ 0, \ (14)(0) \ = \ 0, \ 0 \ = \ 0\)
 
twhitley said:
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
 
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!
 
cool , i got it , thank all of you soooo much, have you looked at any of the others?
 


Oops. I never clicked "Submit". Oh well, I'll post it anyway.

Yes, I'll look at the rest of your exercises later.

twhitley said:
no I'm not adding them , I remember learning to factor and if it added up to the 2nd number then it was right

Oh, okay, now I understand what you're talking about. That four is the coefficient on the x-term in the polynomial.

Yes, you are adding them, heh, heh.

You are correct. When factoring a quadratic polynomial into a product of binomials, you're looking for two numbers that add up to the coefficient on the x-term.


Solving for a, I think of it as solving for x or y, is this correct?

Sure. In the equation a^2 + 4a - 45 = 0, we can think of the variable a as x and 0 as y.

To solve for the equation a^2 + 4a = 45 means to find all Real numbers that when squared and added to four of themselves end up being 45.

 
twhitley said:
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)



If anyone can help, it would be greatly appreciated!

You're given

f(x) = x[sup:3choz3ku]2[/sup:3choz3ku] + 2x - 2

If you want to know what f(3) is, then you'd substitute 3 for each x:

f(x) = x[sup:3choz3ku]2[/sup:3choz3ku] + 2x - 2
f(3) = 3[sup:3choz3ku]2[/sup:3choz3ku] + 2(3) - 2

Then, do the arithmetic on the right side.....
f(3) = (3)(3) + 2(3) - 2
f(3) = 9 + 6 - 2
f(3) = 15 - 2
f(3) = 13

So, f(3) = 13

Try that approach on your problems.
 
twhitley said:
4) sove this system of equations:
5x+4y=-8
2x-y=-11
Multiply the 2nd equation by 4: 8x - 4y = -44

Then add them:
5x + 4y = -8
8x - 4y = -44
===========
13x = -52

Can you finish off?
 
then if I plugged the x value in the equation (-4) then I would get y=3 so would my final answer be x=-4 and y=3
 
I think I would multiply the second equation by 4 so the y would cancel out so I can find value of x, is this correct
 
twhitley said:
3)solve for x: (2/x-1) +1=(2/x^2-x)

If a ratio equals 1, then the numerator and denominator must be the same.

Subtract 2/(x - 1) from both sides.

1 = 2/(x^2 - x) - 2/(x - 1)

Do you know how to combine the two ratios on the righthand side ?

Get a common denominator. Multiply the second ratio by x/x, then both denominators will be the same.

1 = 2/(x^2 - x) - 2x/(x^2 - x)

1 = (2 - 2x)/(x^2 - x)

Since the top and bottom of the righthand side must be the same number, set the numerator equal to the denominator, and solve for x.

Make sure that you check both answers!

 
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