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)

Re: Summer review questions!

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

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.

Re: Summer review questions!

Great, that really help me, thanks. If I have any more questions, I'll post them here.

Re: Summer review questions!

Would:

ln(e^7) = x simplify to 7=x?

Re: Summer review questions!