[QUOTE=Indistinguishable]
We want 3[sup]1/2[/sup] * 3[sup]1/2[/sup] to equal 3[sup]1/2 + 1/2[/sup].
[/QUOTE]
Wait.
If I work that backwards . . .
3[sup]1/2[/sup]+[sup]1/2[/sup] = 3[sup]2/2[/sup] = 3[sup]1[/sup] = 3
So 3[sup]1/2[/sup] = 3
and in ultrafilter’s problem, x[sup]1/2[/sup] = x
Am I right?
[QUOTE=Cisco]
Wait.
If I work that backwards . . .
3[sup]1/2[/sup]+[sup]1/2[/sup] = 3[sup]2/2[/sup] = 3[sup]1[/sup] = 3
[/QUOTE]
This much is correct. The rest is a little off but you’re almost there…
Now, remember, we want 3[sup]1/2[/sup] * 3[sup]1/2[/sup] to equal that quantity 3[sup]1/2[/sup]+[sup]1/2[/sup] which you just showed was equal to 3.
Therefore, we want 3[sup]1/2[/sup] * 3[sup]1/2[/sup] to equal 3.
Rewording that, what we just saw is that we want 3[sup]1/2[/sup], when multiplied by itself, to equal 3.
So what does that make 3[sup]1/2[/sup]? [When multiplied by itself, it should equal 3. Are you familiar with the special name for the operation of multiplying a number by itself?]
[QUOTE=Indistinguishable]
Are you familiar with the special name for the operation of multiplying a number by itself?]
[/QUOTE]
Squaring?
Right. So, if we want the square of 3[sup]1/2[/sup] to be 3, what does that make 3[sup]1/2[/sup]? [Are you familiar with the “opposite” operation to squaring?]
Is it the square root of 3 (which my calculator tells me is 1.73)?
Yes. That is correct.
So, more generally, since x[sup]1/2[/sup] * x[sup]1/2[/sup] should equal x[sup]1/2 + 1/2[/sup], can you carry out all the same reasoning, and tell me what x[sup]1/2[/sup] should be, in terms of x?
Square root of x.
x2 + 3x – 4 = (x + 4)(x – 1) = title changed at the request of the OP.
[QUOTE=SkipMagic]
x2 + 3x – 4 = (x + 4)(x – 1) = title changed at the request of the OP.
[/QUOTE]
Thanks!
[QUOTE=Cisco]
Square root of x.
[/QUOTE]
Correct. Hooray!
<highjack>Indistinguishable, are you a math teacher/professor? You’re really good at this.</highjack>
That lesson was a pain in the ass to learn, but now I don’t think I’ll ever forget that [sup]1/2[/sup] = square root.
What else you got for me?
I know I need a solid base in factoring.
[QUOTE=SkipMagic]
x2 + 3x – 4 = (x + 4)(x – 1) = title changed at the request of the OP.
[/QUOTE]
Aha! Even the mods can’t get the sup tags right!
[QUOTE=Risha]
<highjack>Indistinguishable, are you a math teacher/professor? You’re really good at this.</highjack>
[/QUOTE]
Aw, shucks. blushes No, I’m not a math professor, but I did stay in a Holiday Inn Express last night.
…And I’m a math grad student.
Alright. Are you referring to factoring polynomials or are you referring to something else?
If I were really to complete to exponentiation lesson, I’d have to show you/ensure you grasped how to combine all the things we’ve talked about so far (e.g., by discussing things like “x[sup]-3/2[/sup]”).
One of the questions on the practice test is:
give 6a[sup]2[/sup]b[sup]3[/sup] - 3a[sup]2[/sup]b in factored form
Another is:
For all x is not equal to 0 and y is not equal to 0.
(3x[sup]-2[/sup]y[sup]3[/sup])[sup]2[/sup] / xy = ?
[QUOTE=Cisco]
Another is:
For all x is not equal to 0 and y is not equal to 0.
(3x[sup]-2[/sup]y[sup]3[/sup])[sup]2[/sup] / xy = ?
[/QUOTE]
Ok, let’s tackle this one. Remember that x[sup]a[/sup] * x[sup]b[/sup] = x[sup]a+b[/sup], for all values of x, a, and b. Correspondingly, we always have that x[sup]a[/sup] / x[sup]b[/sup] = x[sup]a-b[/sup], as was basically shown before, simply because division should undo multiplication.
Another useful rule to remember is that (x[sup]m[/sup])[sup]n[/sup] = x[sup]mn[/sup]. To see why this makes sense, suppose you had n rows of xes, each row having m xes (where x is some particular number). The product of all the xes in a row would be x[sup]m[/sup], and the product of that n many times would be (x[sup]m[/sup])[sup]n[/sup]. But that would be the same as the product of all the xes in the entire square, which would be x[sup]mn[/sup]. Make sense?
So, see if you can apply those rules to this problem to try to understand what the numerator comes out to and what the divisions by x and by y do to it.
[QUOTE=Cisco]
One of the questions on the practice test is:
give 6a[sup]2[/sup]b[sup]3[/sup] - 3a[sup]2[/sup]b in factored form
[/QUOTE]
So you have two terms there, each comprising some multiple of 3, some power of a (and a power is kind of a special case of a multiple) and some power of b. Extract the biggest multiple of 3 you can, ditto the biggest powers of a and b, leaving you with something that looks like this:
x(y - z)
where x is the product of the bits you extracted, and y and z is what’s left of your original terms after you’ve divided by the extracted parts.
For instance, if I factorised 5 sin(x)cos[sup]2/sup + 15 sin[sup]2/supcos[sup]3/sup I would get 5 sin(x)cos[sup]2/sup(1 + 3 sin(x)cos(y) )
[sin[sup]2/sup is how (sin(x))[sup]2[/sup] is usually written.]
[QUOTE=Cisco]
One of the questions on the practice test is:
give
in factored form
[/QUOTE]
This question is about knowing how to use the general rule ab + ac = a(b+c). That is, if you can find a common factor in the two terms, you can put them “outside the parentheses”. Here’s a very simple example:
10 = 6 + 4 = **23 + 2*2 = 2 * (3+2) *= 2 * 5 = 10, to demonstrate that the rule works: 10 = 10.
I’ll walk you through the first step in your problem 6a[sup]2[/sup]b[sup]3[/sup] - 3a[sup]2[/sup]b. Knowing that 6=2*3, I see a common factor 3. So put the 3 outside parentheses
Now you can work your way through the remaining 2a[sup]2[/sup]b[sup]3[/sup] - a[sup]2[/sup]b . A hint here would be that b[sup]3[/sup] = b[sup]2[/sup] * b, leading to a common factor b.