SAT Recurring Math Theme?

Test Preparation SAT Preparation
1

<p>Hey, I took two practice SAT tests over the last two days.</p>

<p>In the math portion, I always notice this type of problem.</p>

<p>It has two constants, and you're trying to solve for one. Here's an example of one:</p>

<p>(x-8)(x-k) = x^2-5xk+m</p>

<p>Where k and m are constants. Solve for m.</p>

<p>Another:</p>

<p>t^2-k^2<6
t-k>4</p>

<p>t and k are both positive integers. Solve for t.</p>

<p>I pretty much can't get these types of problems and it drops my math grade by about 50-60 points. Help? :(</p>

2

<p>For the first there’s a basic theorem in algebra – specifically that if the polynomial</p>

<p>x^2 + bx + c = Ax^2 + Bx + C for all values of x, then A = 1, B = b and C = c.</p>

<p>Apply this to the first problem:</p>

<p>Expand (x-8)*(x-k) = x^2 + (-8-k)x + 8k
and this is equal to x^2-5xk+m</p>

<p>Equate coefficients: So 8k = m and 5k = 8 + k. This is easy to solve: k = 2 and m = 16.</p>

<p>You can also solve this intuitively. Try 2 values of x. Start with x = 0, and you get: 8k = m. Then try some other value of x – say x = 8. You get 64-40k+m = 0. Now 40k = 5m. You get m = 16 and k = 2.</p>

<p>For the second problem: t^2-k^2<6
t-k>4. Recall that t^2-k^2 = (t+k)*(t-k) so if (t-k)>4, (t+k)<3/2. I’m not sure that you can solve for t. There’s no doubt more to this problem. But hopefully you get the approach.</p>