<p>I created the following SAT I-like math problem for
your practice. I found it somewhat complicated,
though some may find it very simple:</p>
<p>The function f(x) as shown below (see the tinypic url) is equal to ax^3.
The linear function g(x) intersects the graph of f(x) at
the points (b,2) and (c,-2). The graph of g(x) also intersects the origin.
A line segment, with points D(c,-2) and point E (b,2) is formed with the length of
the square root of 20. What is the value of a? Please
disregard the scale on the x and y axes; the graphs
are not drawn to scale.</p>
<p><a href="http://i30.tinypic.com/9ljyvm.jpg%5B/url%5D">http://i30.tinypic.com/9ljyvm.jpg</a> : Picture of the graph</p>
<p>Tell me if you like the problem and post your answers.</p>
<p>So, the distance from (b,2) to the origin is equal to root(4+b^2). This is eqaul to half of root(20) which is 2root(5)/2=root(5). So, root(4+b^2)=root(5) so 4+b^2=5 so b=±1 (given the y-value, I’m going +1). So, plugging back into the original equation, a*(1^3)=2, so a=2. Yeah? </p>
<p>Anyways terrible SAT problem. Way too much work and values and explanation and stuff. Possibly good SATII problem, though?</p>
<p>1) ab^3 = 2
2) ac^3 = -2</p>
<p>Divide 1) by 2) to get (b/c)^3 = -1 and b/c = -1</p>
<p>sqrt[(b-c)^2 + (2+2)^2] = sqrt (20)
(b-c)^2 + 16 = 20
(b-c)^2 = 4
b-c = 2 (disregarding the negative because a positive minus a negative is always positive)</p>
<p>b/c = -1 –> b = -c</p>
<p>-c-c = 2
-2c = 2
c = -1</p>
<p>ac^3 = -2
a(-1)^3 = -2
-a = -2
a = 2</p>
<p>EDIT: Gamma’s method is much simpler than mine ><. Oh well. Agree that this probably wouldn’t show up on the SAT I (I don’t think I’ve ever had to use the distance formula on it).</p>
<p>I think I mixed way too many concepts into one problem, especially the distance formula, although except the distance formula, this type of problem has came up many times in my SAT I experience.</p>
<p>Here’s an even simpler solution:</p>
<p>ab^3 = 2 and ac^3 = -2</p>
<p>If we assume a, b, and c will be integral answers, then b = 1 and c = -1 (since b = 2 and c = -2 would produce answers larger than 2/smaller than -2). We know b and c are opposites since any y = ax^3 is symmetric with respect to the origin. Then a = 2. Quick application of the distance formula with (1, 2) and (-1, -2) confirms this.</p>
<p>^That works, but I think a and b could also be -.5 and .5, as that gives you a distance between them of 1, which squared is nice. But then you realize that there is no square root of 19. So yeah, this problem can be gamed pretty easily, apparently. Just realize the only squares of integers that add to 20 are 16 and 4 and you’re basically done (I guess that’s another simple method).</p>
