<p>And I will show you how easy it is to solve it</p>
<p>Find all positive integer solutions (x,y) of the equation x^2 + 3 = y(x+2)</p>
<p>Not exactly a realistic SAT question, but intriguing. The answer, I believe, is (5,4) only. I'd love to see a simple, elegant way to get it, if you have one.</p>
<p>(5,4) and
(-1,4)</p>
<p>oh!! only 5,4. i missed the 'positive integers' part</p>
<p>First of all, this is definitely not the type of SAT question you'll see. If you see, it will be multiple question type so you can plug the numbers in and get the answer. Second, I don't even get the point of this thread because it seems like the person wants to show off. Anyhow, here is an explanation for how to solve this problem.</p>
<p>First, divide LHS and RHS (Left Hand Side and Right Hand Side) by x+2. This results in (x^2+3)/(x+2) = y. Using Long Division or Synthetic Division in LHS, you get equal expression as x-2+7/(x+2) = y. Obviously, if x+2 > 7, LHS will include fraction, making y NOT an integer. So, x+2 < 7 and only value that gives integer is when x+2 = 7 or x = 5. Putting 5 in place of x gives y = 4. So, there is one unique solution set of (5,4).</p>
<p>That's a nice way to look at it, grayfalcon.</p>
<p>This was a SAT II (Math II) question in one of my books:</p>
<p>Two roots of x^3 + 3x^2 + Kx - 12 are unequal but have the same absolute value. The value of K is_________</p>
<p>All my rational zero theorm effors have been fruitless, but that is not to say that such an approach could not work.</p>
<p>I don't really understand your question, sl8r000!
the question states that 2 roots are unequal but have the same absolute value...So, let's take a look at this. Given A is a positive number, two roots of A are +a and -a ( with the fact that a^2=A). As you see, +a and -a have the same absolute value: |-a|=|+a|).
Therefore, with your problem, if it doesn't miss anything, I think you just have to find the value of K with which (x^3 + 3x^2 + Kx - 12) is always positive.</p>
<p>We know the sum of the roots must be -3; (-b/a)
We also know the product of the roots must be +12; (- constant term / leading coefficient)</p>
<p>Since two of the roots are opposite in sign but equal in magnitude, the sum of the roots is equal the third root, -3.</p>
<p>The product of the other two roots then be -4, so the other two roots are 2 and -2.</p>
<p>We can FOIL out roots r1, r2, and r3 as</p>
<p>(x-r1)(x-r2)(x-r3) = x^3 + 3x^2 + kx - 12</p>
<p>(x+2)(x-2)(x+3) = x^3 + 3x^2 -4x - 12</p>
<p>so k = -4</p>
<p>Another approach would be to divide x^2 - r1^2 into x^3 + 3x^2 + kx - 12;
Once you realize one factor of 12 must be perfect square, the third root emerges quickly.</p>