# Summer review questions!

• Jul 30th 2011, 03:19 PM
Lotr8808
Summer review questions!
Alright, this summer I have forgotten a lot of what I have learned last year, so I have come to you guys for some help!

I have a few questions, but I'll probably have to add more later.
I want to learn how to do these, not just given the answer, could you break it down step by step?

Here you have to factor and solve for x

$4x^2-3 = 0$

Here, you have to solve for Z:

4x + 10yz = 0

y^2+3yz-8z-4x=0

Thank you :)
• Jul 30th 2011, 03:47 PM
rdtedm
Re: Summer review questions!
1)
a^2-b^2 = (a-b)(a+b)
let a = 2x and b = root(3)
4x^2 - 3 = [(2x-root(3)][2x+root(3)]

[(2x-root(3)][2x+root(3)] = 0 means either 2x-root(3) = 0 or 2x+root(3)=0
so, either x = root(3)/2 or -root(3)/2

2)
4x + 10yz = 0 (subtract 4x from both sides)
10yz = -4x (divide both sides by 10y)
z = -4x/10y (simplify)
z = -2x/5y

3)
y^2+3yz-8z-4x = 0 (subtract y^2 and add 4x to both sides)
3yz - 8z = -y^2 + 4x (factor out z)
z(3y - 8) = -y^2 + 4x (divide by 3y-8)
z = (-y^2 + 4x)/(3y - 8)
• Jul 30th 2011, 04:11 PM
Lotr8808
Re: Summer review questions!
Thank you! I'm just having a bit of a problem on the first one, could you go over it? :)
• Jul 30th 2011, 04:17 PM
rdtedm
Re: Summer review questions!
Quote:

Originally Posted by Lotr8808
Thank you! I'm just having a bit of a problem on the first one, could you go over it? :)

Number one is an example of a difference of squares (a^2 - b^2)
Anytime a problem looks like this, there is a way to factor it. You need to choose a and b so that a^2 and b^2 fit your problem.

Since your problem is: 4x^2 - 3 = 0, we need a^2 to be 4x^2 and b^2 to be 3.
So, a=2x (since (2x)^2 = 4x^2) and b=root(3) (since root(3)^2 = 3)

Once we choose a and b, we know that it factors to (a-b)(a+b). You can confirm this by foiling.. a^2 + ab - ba - b^2 = a^2 - b^2

So, the problem factors to (2x-root(3))(2x+root(3)).

If you have a product of quantities that equals zero, the zero product property states that one or both must be zero. So, to find both solutions of

[(2x-root(3)][2x+root(3)] = 0

We set each term in brackets equal to zero and solve for x.

1) 2x-root(3) = 0 --> 2x = root(3) --> x = root(3)/2
2) 2x+root(3) = 0 --> 2x = -root(3) --> x = -root(3)/2

Hope this makes sense.
• Jul 30th 2011, 04:25 PM
Lotr8808
Re: Summer review questions!
Great, that really help me, thanks. If I have any more questions, I'll post them here.
• Jul 30th 2011, 04:30 PM
Lotr8808
Re: Summer review questions!
Would:
ln(e^7) = x simplify to 7=x?
• Jul 30th 2011, 05:18 PM
rdtedm
Re: Summer review questions!
Yes, it would.