How’s everyone else doing on this quiz? …

How’s everyone else doing on this quiz? …

If you’re on the East Coast, I hope that you and yours are able to weather the storm OK. I’m not sure whether I’ll have power or not on Monday, but I’ll try to get the solution posted then if possible.

This week’s prize will again be access to the Math Guide Beta, which is looking radder and radder by the day. I’ve been doing a TON of work on it. Anyhoo, first correct (and *not anonymous*) answer in the comments gets access.

If (

x^{2}+ 5x)^{2}– 36 = (x+m)(x+n)(x+p)(x+q), what is the median of the set {m,n,p,q}?

Good luck, kiddos. And stay safe.

UPDATE: Nice work, Ammad. I hope you enjoy the book, which I’ve now shared with you in Google Docs.

Solution below the cut.

The trick to getting this one right (or at least, to getting it right without the aid of a powerful calculator) is recognizing that the left side of the original equation *is actually the difference of two squares*:

(*x*^{2} + 5*x*)^{2} – 36

(*x*^{2} + 5*x* + 6)(*x*^{2} + 5*x* – 6)

Of course, each of those can be factored as well:

(*x* + 3)(*x* + 2)(*x* + 6)(*x* – 1)

So although we don’t know which one is which, we *do* know that {m, n, p, q} = {-1, 2, 3, 6}, which has a median of 2.5.

As usual, this was a bit tougher than you’d see on the SAT, but if you figured out that, despite its complexity, this was really just a difference of two squares question, then you deserve a cookie. No…*two* cookies.

Very sorry for the disappearing and reappearing blog the past 24 hours or so. Blogger has had some difficulties, but I’ve used it for the better part of a decade without incident, so this one won’t send me running for the hills. I’m assured that the super long post I wrote and posted yesterday about obsessive vocabulary studying will be back shortly. Or I will seriously freak out.

Prize for answering this weekend’s challenge question: I’ll use an image of your choice in a future post. Image can’t be copyrighted, profane, or have anything to do with the Philadelphia Phillies. I reserve the right not to post anything I think sucks.

Your weekend challenge:

If

pandqare positive integers, what isp+q?

Put your answers in the comments; I’ll post the solution Monday. Good luck!

UPDATE: Congrats to JD for getting it the long way, and then the short way. Solution below the cut.

It’s funny: I write these questions in a vacuum and sometimes I don’t realize exactly how hard they’ll be until I discuss them with a few people. This one, as these challenge questions go, was especially difficult. I knew it was devious to use 243 in the exponent and the denominator, since the exponent literally could have been anything, but I just couldn’t help myself. So I’m sorry if this one drove you nuts.

There are two insights required to solve this:

**15**. This is easier to see when you’re dealing with variables (ex: (^{243}= 3^{243}5^{243}*xy*)^{2}=*x*^{2}*y*^{2}), but you can always factor numbers that are raised to exponents, if that’ll help you towards a solution. In this case we’re trying to get to 3^{p}5^{q}; that’s a clue that you’re going to want to factor the 15 out.**243 = 3**. Yeah. That’s gonna be important.^{5}

Given those things, let’s attack the problem (remember, every step you take should bring you closer, somehow, to 3^{p}5^{q}.

Now you can cancel the 3^{5} out of the denominator by subtracting 5 from the exponent in 3^{243} in the numerator. (For a review of this and other exponent rules, click here.)

And there you have it. If *p* and *q* are positive integers, then they have to be 238 and 243, respectively. So *p* + *q* = 481.

If I asked told you it was my birthday and I wanted a cake, what would you do? You’ve got two choices: buy a bunch of ingredients and start baking, or go to a different aisle in the grocery store and just buy the cake.

Baking the cake yourself is not only more time consuming than just buying one; it also gives you more opportunities to screw up (like, say, mistake salt for sugar and bake the grossest cake of all time). Since you know I’m a shameless crybaby who will never let you forget it if you ruin my birthday, you should just buy the cake in the cake aisle, and then use your time to do something more fun than baking.

And so it is with the SAT. What do I mean? WHAT DO I MEAN??? Read on, young squire.

Here’s a pretty common question type on the SAT:

- If 3
x–y= 17 and 2x– 2y= 6, what is the value ofx+y?

(A) 8

(B) 9

(C) 11

(D) 12

(E) 14

The SAT is asking you for a cake here. Baking it yourself will still result in a cake, but it will also give you opportunity to screw up, and take longer than just buying one. They don’t give a rat-turd if you buy the ingredients (*x* and *y*), so **don’t waste time on them**! All that matters is the finished cake, (the value of the expression *x* + *y*), and we should be able to solve for that directly without ever finding *x* or *y* individually.

To do so, first stack up the equations we’re given and the expression we want:

3*x* – *y* = 17

2*x* – 2*y* = 6

*x* + *y* = ?

Do you see it yet? How about now:

3*x* – *y* = 17

–[2*x* – 2*y* = 6]

*x* + *y* = ?

That’s right. All we need to do is subtract one equation from the other (I’ve already distributed the negative here):

3*x* – *y* = 17

-2*x* + 2*y* = -6

*x* + *y* = 11

Our answer is (C) 11.

Note that we didn’t have to do any algebra here (aside from distributing that negative). This is not the exception on the SAT, this is the rule. When you’re given two equations and asked to solve for an expression, you almost NEVER have to do algebra. So instead of jumping into an algebraic quagmire as soon as you see questions like this, ask yourself “*How do I quickly go from what they gave me to what they want?*” Let’s look at one more example together:

- If (
x–y)^{2}= 25 andxy= 10, then what isx^{2}+y^{2}?(Let’s call this one a grid-in, so no multiple choice.)

Here, we aren’t going to get anywhere by simply adding or subtracting our two equations, but do you see anything else going on? Do you see that you’ve been provided with all the pieces of a particular puzzle?Let’s start by expanding what we were given (FOIL works here of course, but you should really make an effort to memorize the basic binomial squares and difference of two squares to save you time on the test; they all show up frequently):

(*x* – *y*)^{2} = 25

* x^{2} – 2xy* + y

And would you look at that—we’re pretty much already done. Just substitute the value you were given for *xy*, and do a little subtraction:

* x^{2} – 2(10)* +

Note again that we didn’t need to solve for *x* or *y* individually to find this solution; all we needed to do was move some puzzle pieces around. Note also that actually solving for the individual variables would have been a *huge* pain.

Again, your mantra: “*How do I quickly go from what they gave me to what they want?*“

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Also try this question I posted a while back (contains explanation).

- For real numbers
a,b, andc,ab= 1.5,bc= 6, andac= 25. Soabc=(A) 9

(B) 12

(C) 15

(D) 100

(E) 225

Answer and explanation below the fold…

If I told you it was my birthday and I wanted a cake, and you liked me enough to provide one (pretty please?), you’d have a couple options. You could go to the grocery store and buy all the ingredients, and then go home and start baking. Or, you could go to the same store and go to a different aisle, buy a pre-made cake, and spend the time you would have been baking watching awesome videos on the internet. Listen, I’m glad you like me and all, but you don’t really need to take all that time making me a cake. Internet videos. Get some.

What I mean to say is: when the SAT asks you to find an expression (by which I mean anything other than just a plain old variable like *x*) you should look for a way to **solve directly for that expression**. Your mantra for a question like this: “** How do I go from what they gave me to what they want?**”

Often, you’ll be adding or subtracting two equations. That doesn’t do us any good with this problem, though. For this one, we’re going to have to get a little more creative. Note that we are given three equations containing parts of what we want: *abc*. How do we go from what they gave us to what they want?

Try multiplying everything you have together:

(*ab*)(*bc*)(*ac*) = (1.5)(6)(25)

*abbcac* = 225

Now simplify a little bit:

*aabbcc* = 225

*a*^{2}*b*^{2}*c*^{2} = 225

(*abc*)^{2} = 225

And take the square root of both sides:

*abc* = 15.

That’s answer choice (C).