Equation ManipulationDate: 4/13/96 at 20:3:29 From: Anny Subject: SAT problem Hi. I'm taking the Princeton Review Course right now and my instructor and I are having a problem with a certain question. We know the answer, but just can't figure out how it works. So, here goes... Number 10 on Section 6 of a practice SAT: If m and n are integers, which of the following could be equal to {mr + ns}^2 for all values of r and s? (A) 2r^2 + 4rs + 2s^2 (B) r^2 + 4rs + s^2 (C) r^2 + 2rs + 4s^2 (D) r^2 + 2rs - s^2 (E) r^2 - 2rs + s^2 If you could help out it would be greatly appreciated. I hope you guys have better luck than we did! Date: 4/14/96 at 15:9:31 From: Doctor Syd Subject: Re: SAT problem Hello! Let's first write out (mr + ns)^2: (mr + ns)^2 = m^2r^2 + 2mnrs + n^2s^2 Right? Okay, So, this means the coefficient of r^2 must be a square of an integer, right? Well, this rules out part (a) since 2 isn't the square of an integer. For b through e, suppose each of the given answers could be equal to (mr + ns)^2, and figure out what the values of m and n would then have to be. You'll get a contradiction for all but one of them. For instance, let's do (b): If r^2 + 4rs + s^2 = m^2r^2 + 2mnrs + n^2s^2 then we must have that m^2 = 1 and n^2 =1, and 2mn = 4. But this can't be since m^2 = 1 and n^2 = 1 implies that m is plus or minus 1 and n is plus or minus 1, so mn is plus or minus 1 (considering all of the cases), so 2mn is plus or minus 2 which is not equal to 4. See if you can try out c-e and figure out which one doesn't give you a contradiction! Hope this helps. Write back if you have more questions. -Doctor Syd, The Math Forum |
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