# I have a month to learn college algebra **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [June 21, 2008, 2:17am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656 "2008-06-21T02:17:45Z")\ **Posts on this page:** 18\ **Page:** 6
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 3:55am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/101 "2008-06-27T03:55:10Z")
[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? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 4:06am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/102 "2008-06-27T04:06:36Z")
[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?] ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 4:13am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/103 "2008-06-27T04:13:21Z")
[QUOTE=Indistinguishable] Are you familiar with the special name for the operation of multiplying a number by itself?] [/QUOTE] Squaring? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 4:15am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/104 "2008-06-27T04:15:23Z")
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?] ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 4:18am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/105 "2008-06-27T04:18:06Z")
Is it the square root of 3 (which my calculator tells me is 1.73)? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 4:25am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/106 "2008-06-27T04:25:24Z")
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? ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 4:27am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/107 "2008-06-27T04:27:29Z")
Square root of x. ---
**Author:** ![SkipMagic](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skipmagic/32/20706_2.png) [@SkipMagic](https://boards.straightdope.com/u/SkipMagic)\ **Post date:** [June 27, 2008, 4:34am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/108 "2008-06-27T04:34:50Z")
x2 + 3x – 4 = (x + 4)(x – 1) = title changed at the request of the OP. ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 4:36am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/109 "2008-06-27T04:36:55Z")
[QUOTE=SkipMagic] x2 + 3x – 4 = (x + 4)(x – 1) = title changed at the request of the OP. [/QUOTE] Thanks! ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 4:38am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/110 "2008-06-27T04:38:11Z")
[QUOTE=Cisco] Square root of x. [/QUOTE] Correct. Hooray! ---
**Author:** ![Risha](https://avatars.discourse-cdn.com/v4/letter/r/5fc32e/32.png) [@Risha](https://boards.straightdope.com/u/Risha)\ **Post date:** [June 27, 2008, 4:42am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/111 "2008-06-27T04:42:36Z")
\ **Indistinguishable** , are you a math teacher/professor? You’re really good at this.\ ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 4:51am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/112 "2008-06-27T04:51:20Z")
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. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 27, 2008, 5:19am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/113 "2008-06-27T05:19:31Z")
[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] \ **Indistinguishable** , are you a math teacher/professor? You’re really good at this.\ [/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. > [@](#): > > What else you got for me? > > I know I need a solid base in factoring. 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]”). ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 6:04am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/114 "2008-06-27T06:04:18Z")
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 ---
**Author:** ![Cisco](https://avatars.discourse-cdn.com/v4/letter/c/ccd318/32.png) [@Cisco](https://boards.straightdope.com/u/Cisco)\ **Post date:** [June 27, 2008, 6:06am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/115 "2008-06-27T06:06:20Z")
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 = ? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 28, 2008, 2:45am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/116 "2008-06-28T02:45:58Z")
[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]m_n[/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]m_n[/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. ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [June 28, 2008, 9:11am UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/117 "2008-06-28T09:11:12Z")
[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.] ---
**Author:** ![4.66](https://avatars.discourse-cdn.com/v4/letter/4/2bfe46/32.png) [@4.66](https://boards.straightdope.com/u/4.66)\ **Post date:** [June 28, 2008, 1:05pm UTC](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656/118 "2008-06-28T13:05:05Z")
[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] 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 = \*\*2_3 + 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 > [@](#): > > _6a[sup]2[/sup]b[sup]3[/sup] - 3a[sup]2[/sup]b_ =_3_2a[sup]2[/sup]b[sup]3[/sup] - 3a[sup]2[/sup]b\* = _3(2a[sup]2[/sup]b[sup]3[/sup] - a[sup]2[/sup]b)_ 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_. [Previous page](https://boards.straightdope.com/t/i-have-a-month-to-learn-college-algebra/453656.md?page=5)