<p>This problem is from the SAT II Level 1 Math sample question pdf on the collegeboard site.</p>
<p>f(x) = x^4-3x^3-9x^2+4</p>
<p>How many real numbers K does f(k) = 2?</p>
<p>I subtracted 2 from both sides and got the equation: x^4-3x^3-9x^2+2 = 0</p>
<p>This is nonfactorable by grouping, does not provide any roots, and has no quadratic form.</p>
<p>Is there a fast way to do this?</p>
<p>Would I have to do rational root test and plug them all into synthetic division?</p>
<p>Couldn't you graph it on your calculator?</p>
<p>Yeah you're going about it the hard way. :D Graph baby, graph!</p>
<p>or you could look at the answer choices</p>
<p>Or get TI's own polynomial root finder for the 83+...</p>
<p>i just graphed it, problem took me 10 seconds</p>
<p>well if you wanted to do it the hard way...</p>
<p>we could do:</p>
<p>x^2(x^2 - 3x - 9) = -2</p>
<p>we can void the x^2 root(s) since x^2 is greater or equal to zero. then we just need to know:</p>
<p>x^2 - 3x - 9 = -2</p>
<p>X^2 - 3x - 7 = 0</p>
<p>[ 3 +/- sqrt(37) ] / 2</p>
<p>I believe that is correct at least...but did it kinda quickly just for my own amusement...</p>
<p>Approach 1: Graph.
Approach 2: Synthetic division/rational roots thm.
Approach 3: Program that solves for the roots.</p>
<p>wow that was really dumb of me ignore that. this is the same brain that seems to have madly deteriorated from my 800 math in june to the sub-700 math for nov. lol</p>
<p>Just a quick question:</p>
<p>finding zeros is the same thing as finding roots or does one refer to only real x-intercepts while the other refers to both real and imaginary? Or does it not make a difference?</p>
<p>Thanks!</p>
<p>stambliark, why did you just get rid of the X^2 term like that?</p>
<p>Also, I tried rational roots theorem, and none of the rational roots are roots of the equation.</p>
<p>Il Bandito:
This is not a terribly scientific way to go about doing it, and it may happen to work quickly just for this specific quartic function, but here goes:</p>
<ol>
<li><p>Define g(x) = x^4-3x^3-9x^2+2. You need to find out how many real roots this equation has.</p>
<ol>
<li>Since you have (1)x^4 as part of the equation, this will dominate for very large +ve or -ve values of x; so the curve goes soaring towards +infinity when you get far away from the origin, on both sides of the y-axis.</li>
<li>Try simple values of x=0,1,-1; you get
g(0) = 2
g(1) = 1 -3 - 9 + 2 = -9
g(-1) = 1 +3 -9 + 2 = -3</li>
</ol></li>
</ol>
<p>So from a value of 2 at x=0, the curve g(x) crosses the x-axis (ROOT1) to get to a value of -9 at x=1, and must re-cross the x-axis (ROOT2) to eventually start going towards +infinity, as you keep increasing the value of x. Similarly, from a value of 2 at x=0, the curve g(x) crosses the x-axis (ROOT3) to get to a value of -3 at x=-1, and must re-cross the x-axis (ROOT4) to eventually start going towards +infinity, as you keep decreasing the value of x.</p>
<p>There are 4 real numbers for which g(x) = 0. Good thing they didn't ask you to find the actual values of the 4 roots, though.</p>