<p>Which of the following gives all values of k for which x^3 - 3x^2 -9x = k has 2 or more real roots?</p>
<p>(A) k> 27 or k < -5
(B) -5 <= k <= 27
(C) -27 <= k <= 5
(D) k = 5 or k = -27
(E) -27 < k < 5</p>
<p>I have the answer, but I would like an explanation. Thanks! :)</p>
<p>By the way, this is from Meylani's book.</p>
<p>Also, how does one go about doing the following:</p>
<p>x + yi = 1.8 - 2.4i
–> x - yi = 1.8 + 2.4i</p>
<p>How does this step take place? It’s in the explanation of the book I’m using.</p>
<p>(x+iy)(x-yi) = x^2 + y^2</p>
<p>(1.8 - 2.4i)(x - yi) = 1.8^2 + 2.4^2</p>
<p>And go from there.</p>
<p>For the other one,</p>
<p>Just factor out the x; x(x^2 - 3x -9) = k</p>
<p>And use the rule for two distinct roots: b^2 - 4ac > 0</p>
<p>@jsanchez: Thanks, I get the imaginary number one, but the other one I tried your method and am not getting the right answer. :/</p>
<p>x(x^2 - 3x - 9) = k</p>
<p>a = 1, b = -3, c = -9</p>
<p>b^2 - 4ac > k</p>
<p>(-3)^2 - 4(1)(-9) > k</p>
<p>9 + 36 > k</p>
<p>k < 45</p>
<p>Am I doing this wrong? :(</p>
<p>@jsanche32 You’re logic is correct, however, this only works with a polynomial of degree 2, or a quadratic. Because this function is a cubic, solving for the determinant wont work.</p>
<p>@eobaggs. So, what I would do is first rewrite the equation as y= x^3 -3x^2 - 9x -k. Now, just graph the function and ignore k, so let k=0. When you graph it, you should see a weird function that has a local maximum at (-1, 5) and a local minimum at (3,-27). Since the k value will move the function up or down, if it moves the function more than 5 down, then the function will only have 1 real root, and if it moves the function more than 27 up, then the same thing will happen. </p>
<p>So, in mathematical terms, k <= -27 or k => 5, so the answer is C.</p>
<p>Is that correct/helpful?</p>
<p>Ah, thanks so much EducationOD! That makes sense! However, the answer is C, meaning that K can be 5 or -27, which makes sense I think. That was very helpful. Nice of you to even graph it. 
</p>
<p>Here’s another problem I’ve ran into some trouble with:</p>
<p>What does the equation 2x^2 - x + 2y^2 + y = 0 represent?</p>
<p>a) circle
b) ellipse
c) parabola
d) hyperbola
e) none of the above</p>
<p>I cannot seem to get the equation down to any simpler form. :(</p>
<p>Here’s what I would do.</p>
<p>First move k over, so you’ve got xxx-3xx-9x-k=0. If you think about it this way, k can move the graph of xxx-3xx-9x up or down. Graph xxx-3xx-9x on your calculator and find the relative maximum (5) and minimum (-27). To keep at least two real zeroes, then, you can move the graph down by no more than 5 and up by no more than 27. Because you made k negative, this means that -27<=k<=5.</p>
<p>EDIT: Whoops, too late! EducationOD has it backwards, though, because k is negative in the rearranged equation.</p>
<p>2x^2 - x + 2y^2 + y = 0</p>
<p>You see:
2 squared terms, same sign, same coefficient.</p>
<p>Circle. Done.</p>
<p>To get the center you would have to complete the squares and factor though.</p>
<p>@eobaggs… lol that’s what I said… </p>
<p>And ok, this next one is actually tricky. So what you want to do is complete the square. I’m assuming you don’t know how to do this, but you probably do, and if so, then you can skip to the next paragraph. So completing the square basically makes the equation seem prettier, and easier to understand. First, you can divide the whole right side by 2, and get x^2 - x/2 + y^2 +y/2. </p>
<p>Now, you can add both 1/16 to both sides to get (x-1/4)^2 +y^2 +y/2 = 1/16. Again add 1/16 to both sides to get (x-1/4)^2 + (y+1/4)^2 = 1/8. Now you can multiply both sides by 8 to get 8(x-1/4)^2 + 8(y+1/4)^2 = 1, which is the equation for a circle. </p>
<p>Alternatively, the way I would look at it is this.
Since you have both an x^2 and a y^2 term, the equation will be either a circle, ellipse, or hyperbola. Since you’re ADDING x^2 and y^2, then your equation is ether a circle or ellipse, and finally, since the coefficients for x^2 and y^2 are equal (both are 2), it must be a circle.</p>
<p>@Daisie: Ah, thanks for clearing that up! </p>
<p>@OtherWindow: Ah, that’s an easy way to think of it! Thanks, ha. I’m going to have to review my completing the squares, completely forgot about that, although I doubt that’ll be that important on the test.</p>
<p>@EducationOD: Thanks so much for taking the time to help me, first off! </p>
<p>And thanks, ha, I completely forgot how to complete the square. Now, however, how do you know to add 1/16? Is there like a shortcut?</p>
<p>Nevermind, I found this: [Completing</a> the Square: Solving Quadratic Equations](<a href=“http://www.purplemath.com/modules/sqrquad.htm]Completing”>Completing the Square: Solving Quadratic Equations | Purplemath)</p>
<p>Thanks everyone! :)</p>
<p>I assumed you weren’t allowed to use graphing calculators when I posted my response.</p>
<p>So you do it via inspection.</p>
<p>Just factor out the x; x(x^2 - 3x -9) = k</p>
<p>The quadratic in brackets has b^2 - 4ac > 0 (so Ok)</p>
<p>Then let f(x) = x^3 - 3x^2 -9x - k<br>
So
f(-2) = -2-k
f(-1) = 5-k local max at x= -1 Here k = 5 for f(-1) = 0
f(0) = -k<br>
f(1) = -11 - k
f(2) = -22- k
f(3) = -27-k (Local min at x = -3) Here k = -27 for f(3) = 0
f(4) = - 20-k</p>
<p>Answer: -27<= k<=5</p>