<p>Can someone explain how to solve this problem?
1. If a and b are positive integers and (a^6b^4)^1/2 = 675, what is the value of a+b?
Thanks!</p>
<p>The left side is a^3b^2. And the right side factors: it’s 27x25…</p>
<p>To clarify on the above post, taking the square root of the left side (a^6 * b^4) gives a^3 * b^2. Then, if you prime factorize the right side (675), you get 3^3 * 5^2. This lines up quite nicely with the expression on the left side. Thus, a = 3 and b = 5 and a + b = 8.</p>
<p>^ Agreed. And I like the phrase: “this lines up quite nicely”. That’s the thing about SAT problems. They are designed to be solved and there are no coincidences. So as soon as you see (a^3)(b^2) on the left you have to think “Why would they set that equal to 675? 675 is such a random number”. But then you factor it and see that it is not random at all. It was chosen to fit the problem. The stuff they give you and the stuff they ask for are related :)</p>
<p>@pckeller So it would be unlikely for a question to require a lengthy solution?</p>
<p>Even the hardest SAT problems can be solved in under a minute and in no more than say 3 steps. I’m not saying that every student will find the quick path, but it always exists. You really have to do lots of practice to develop your own feel for things. I tell my students that if they see a long path and they feel comfortable with the steps, go ahead and follow it. But often it is worth investing a moment or two to try to find the shorter path.</p>
<p>That is also why you should be spending lots of time reviewing your blue book tests – even the problems you got right! Go back and see how you could have attacked the problems differently. The more you practice with alternate approaches, the more comfortable you get with them.</p>
<p>And look closely at what is given and what is asked for. For example:</p>
<p>Given (x+y)^2 = 400 and (x-y)^2 = 100 find 8xy.</p>
<p>@pckeller im just gonna do a little quick solution to that problem you posed.
(x+y)^2 = 400 and (x-y)^2 = 100
x^2+2xy+y^2=400 x^2-2xy+y^2=100<br>
x^2+2xy+y^2 - (x^2-2xy+y^2) =400 -100
4xy = 300
therefore 8xy = 600.
there’s no easier solution Im missing is there?
(Im going into college but Im a math nerd) </p>
<p>@guineagirl96 that’s not a bad solution; it should take no more than 30 seconds if you’re familiar with binomial theorem or “foil.”</p>
<p>Alternatively, let x = 15 and y = 5. Then 8xy = 8<em>15</em>5 = 600.</p>
<p>^Wow. That’s an insightful approach.</p>
<p>^, ^^, ^^^ Yes to all! Also notice that when I wrote the problem, I intentionally made it so that finding numbers would not be that hard. But here’s something I have been noticing lately: the students who can do things by playing around with numbers are often the ones who know their algebra backwards and forwards. Let me explain…</p>
<p>In school, you are taught how to translate word problems into equations. You see “John has 3 more than twice as many marbles as Mary” and you automatically (I hope) think: J= 3 + 2M. That’s “forwards”. But I am not sure that as many students can do the opposite, “backwards” process as automatically. When you see: (x+y)^2 = 400 and (x-y)^2=100, can you translate that back into a word problem? Think: OK, there are two numbers. When I add them and square them, its 400 so a sum of 20 would work. But when I subtract and square, its 100 so a difference of 10 would work. Now what two numbers add up to 20 but subtract to 10? Hmm…</p>
<p>(I bet MITer94 did all of that in about 12 seconds.)</p>
<p>It is probably said above, but there is nothing arbitrary in “Alternatively, let x = 15 and y = 5” </p>
<p>That is NOT a plugin in the strict sense, but a direct observation of </p>
<p>(x+y)^2 = 400 and (x-y)^2 = 100
or
(x+y) = 20 and (x-y) = 10
or
x+y+x-y = 20 + 10
or 2x = 30 </p>
<p>This is done without having to write the equation as the numbers are so easy to manipulate to obtain 15 and 5. </p>
<p>Would this one have been more interesting? </p>
<p>Given (x+y)^2 = 400 and (x^2-y^2)= 200 find 8xy.</p>
<p>@xiggi i don’t SAT would ask that really that would involve using the quadratic equation to solve and involves square roots, plus has multiple correct answers.</p>
<p>It is more interesting! But I still get 600…</p>
<p>Now we are looking for a pair of numbers that add up to 20 (or negative 20) and if you factor out both expressions and divide one by the other, you find that (x+y)/(x-y)=2. But we already knew that their sum was 20 so that means their difference is 10. And we are back where we started.</p>
<p>oh so (x+y)^2/[(x+y)(x-y)]=400/200 x+y=20
(x+y)/(x-y)= 2
let z= x-y
20/z =2
therefore x-y=10
so you get 15, 5
which gives you the 600… didn’t see that before duh! I forgot (x^2-y^2) factored… oops not a good sign as a perspective math major coming in taking linear algebra, but i’ve made worse mistakes XD</p>
<p>Here is a blog post which shows another way to picture those factors:</p>
<p><a href=“http://advancedmathyoungstudents.com/blog/2014/03/27/it-really-is-the-difference/”>http://advancedmathyoungstudents.com/blog/2014/03/27/it-really-is-the-difference/</a></p>
<p>^ Slightly more interesting, indeed, but not too much harder. </p>
<p>@guineagirl96 yeah I’ve made many many mistakes too. I had an AIME once where I wrote a 3 in the exponent instead of 2.</p>