<p>Do you mind expanding the left side of the equation?</p>
<p>We don’t know what k is yet, so here:
2x^2 + (b-14)x - 7b = 2x^2 - 11x + k
…Can’t you please just tell me how to get from that to b-14 = -11 and to -7b = k.</p>
<p>We don’t need b here, because in the test, you want to do it quickly.
Ok, we have</p>
<p>(x - 7)(2x + k/-7) </p>
<p>= 2x^2 + k/(-7) * x - 14x -7*k/(-7)</p>
<p>= 2x^2 + [k/(-7) - 14] x + k (1)</p>
<p>this should equal to </p>
<p>= 2x^2-11x + k (2)</p>
<p>compare (1) and (2)</p>
<p>we have k/-7 -14 = -11</p>
<p>so k/-7 = 3</p>
<pre><code>k = -21
</code></pre>
<p>see, we don’t need b. We only need to know how to do the expansion. </p>
<p>I can tell you that you will not see anyone else do the problem like this, this is way too fast because you don’t need a pencil and paper. I can do it all in my head. I learned this from a Faster-Than-Light Math! tutor with a Ph.D. degree.</p>
<p>Uh, how do you know even know what B is? How did you get rid of it?
:/</p>
<p>Can someone please just tell me how to get from 2x^2 + (b-14)x - 7b = 2x^2 - 11x + k to b-14 = -11 and to -7b = k??? Test is tomorrow! D:</p>
<p>2x^2 + (b-14)x - 7b = 2x^2 - 11x + k </p>
<p>so (b-14)x - 7b = - 11x + k </p>
<p>so (b-14) = -11 for the x</p>
<p>and -7b = k</p>
<p>but this is way too slow to do it. You really have to visualize the process that I mentioned before. Here is the Youtube video that shows you how to visualize those algebra functions:</p>
<p>[YouTube</a> - SAT Math Algebra Function Problem](<a href=“- YouTube”>- YouTube)</p>
<p>Okay, but here’s my main problem!! Haha, WHY (b-14) = -11 and why does -7b=k? Is that some kind of property?</p>
<p>(x-7)(2x+a)</p>
<p>We use “a” just to represent the missing variable</p>
<p>When we expand…
2x^2 + ax -14x - 7a
But the real equation is
2x^2 -11x + k</p>
<p>-7a = k, but we can’t do anything there yet. </p>
<p>However, we see that ax-14x = -11x<br>
So a must = 3. Right?</p>
<p>Now we can plug “a” into the other equation we couldn’t do before</p>
<p>-7(3)= -21</p>
<p>Thus, k=-21</p>
<p>But how did you get -7a=k? Sorry and thanks. :/</p>
<p>K is a constant. The only way to get constant K is to multiply g and h in the expression: (x-g)(x-h).</p>
<p>Huh…I guess I understand. But this just seems out of the ordinary. Thanks for the help RAlec and easiermath! Although I still hope this doesn’t show up tomorrow, I might be able to do it if I see it! Thank you so much!</p>
<p>Turbo64, you are welcome. You may get the fish, but I don’t think you know how to fish yet. In other words, solving the problem without introducing any variables, without a pencil and paper, alll in 30 seconds or less. Moreover, to use the methods to solve other problems in the similar fashion. If you need more explaination or summary, I will help you.</p>
<p>@ zslavitz</p>
<p>Hmm, how did you go about getting 8? Ah, I think I see, unfortunately it’s not quite as simple as just 60 - 52 = 8. The question seems to be asking what distance on that stretch of highway that both radio stations are heard, not (a much more complicated) question of what is the overlapping area between the two radio stations… but that sounds like it would be fun to do…</p>
<p>In any case if you draw a diagram:
A - Station 1
B - Station 2
=== - Highway</p>
<p>…52 miles … 52 miles …48 miles…60 miles…
|-----------|-----------|---------|-------------|
=========A================B===========
|-----------|------|-------------|--------------|
… 52 miles…40 miles…60 miles … 60 miles …</p>
<p>Erm, right, so the distance between A and B MUST be 100 miles. So above the “highway” you can see the full representation of the radius of station A and below the “highway” you can see the full representation of the radius of station B. Unfortunately you can’t effectively fully describe the radii of both with text on here… without color or something… this is due to the overlap.
…52…48…
So if you notice that on top of the highway it goes |-----------|---------|
…40…60…
and below it is…|-------|-------------|
…|—|…
so if you look at the overlap between the 52 and the 60 the space is 12 miles.
Arithmetically this is simply 52 - 40, or 60 - 48. Visually this is much easier to see than describe.</p>
<p>Hope I helped :).</p>
<p>@ Turbo</p>
<p>What most people have been replying with are complete walkthroughs of the problem. And really the method’s presented are quite effective and will be an invaluable tool in some of the hardest problems on the SAT/ACT. However to answer simply what you asked in the initial post I trued covering just that bit below (since it seems whatever source you were using partially walked you through it already, you just didn’t understand how it made one jump in the steps). I went veeeeeeeery detailed and probably included patronizingly basic steps but I wasn’t sure exactly where you had issue so I went all out. While I hope I hit what you were looking for, I can’t be sure.</p>
<p>So, at the point where you have:</p>
<p>“2x^2 + (b-14)x - 7b = 2x^2 - 11x + k
THEN, somehow, they get b-14 = -11 and that -7b = k. Can anyone explain how they get that?”</p>
<p>The simplest way to think about this in terms of substitution. If you look at the left and right sides of the equation the difference is that in the original equation (on the right) the coefficient of x is -11 and the final constant being added on the end is k. While on the derived equation (on the left) the coefficient of x is (b-14) and the constant being added on the end is -7b. Simply taking this at face value it logically makes sense that if you can substitute one for the other they must be the same. This will hold true in all cases and is simply a part of mathematical logic.
This, as far as I know, doesn’t have an explicit property devoted to it. It can probably be derived from more basic properties but it honestly isn’t that difficult of a conceptual hurdle and is done quite a bit in more in some crazy advanced mathematics.</p>
<p>Of course you could “try” to simply things down… sort of… before doing that.
.2x^2 + (b-14)x - 7b = 2x^2 - 11x + k
-2x^2…-2x^2
So you subtract 2x^2 from both sides of the equation.
(b-14)x - 7b = -11x + k
At that point it may be easier to see the potential for substitution.</p>
<p>The way you taking the problem, it relies on you equating (b-14) = -11 and -7b = k.
This, of course, isn’t the final solution and there are further steps to take. So, really, you could take a slight detour if you don’t like that step and go about it (slightly) differently.
We already got here:
(b-14)x - 7b = -11x + k</p>
<p>At this point you could rearrange a bit:
(b-14)x - 7b = -11x + k</p>
<h2>. + 11x + 7b = +11x + 7b</h2>
<p>(b-14)x + 11x = k + 7b</p>
<p>Than you can distribute the x and combine terms:
bx - 14x + 11x = k + 7b
bx - 3x = 7b + k</p>
<p>From here you have to rely on the standard method of obtaining the “roots” or “zeroes” of a function. Each side of the equation is equal to 0 (this also makes sense if you consider the (b-14) = -11 and -7b = k again). </p>
<p>bx - 3x = 0
The only value for be that would hold true for all x values would be 3. You could also +3x to both sides of the equation then divide both sides by x as seen below:
bx - 3x + 3x = 0 + 3x
bx = 3x
bx/x = 3x/x
b = 3</p>
<p>We also see that k + 7b = 0
k + 7(3) = k + 21 = 0
k = -21</p>
<p>That’s basically how it all plays out. It’s not that difficult of a problem, you just can’t let yourself be caught up by some of the conceptual leaps. Math works, you just have to keep faith in that.</p>
<p>As for 58, it’s hard to say without seeing the actual image in question but I can try helping (it’ll be much more brief, trust me).
“What is the measure of <COD measured in the direction of the arrow shown?” The question you’re referencing is asking what the angle is of COD (that is, starting at point C if you trace the line connecting it to O and from there to D what is the angle it makes right up against point O?).</p>
<p>The rest of the statement is merely technicality math jargon, but isn’t entirely necessary. Without any further information I’m going to assume the angle is in standard position (one side or “leg” of the angle is along the positive x-axis. So I’m going take a stab at sketching:</p>
<p>…^
…/
…D
…/
…/
…/
…/.))
O------------C------></p>
<p>So you want the angle measure ( represented by the two close parentheses “))” ).
Of course the angle probably looks different but the “direction of the arrow show” would be the final “leg” of the angle, in this case the leg comprised of O–D. This should be simple enough but may be more difficult if they give you limited information.</p>
<p>Hope at least some part of this helped :).</p>