<p>Two solutions of x^2-9x+c=0 are complex conjugate, which of the following describes all possible values of C</p>
<p>1.c=0
2.c=is not zero
3.c>81/4
4.c>81
5.c>9</p>
<p>Can someone explain the reason behind this? thank you so much!</p>
<p>A quadratic with real coefficients either has two real roots, one double root, or two non-real roots which are complex conjugates. Therefore the determinant is less than zero, that is,</p>
<p>(-9)^2 - 4(1)(c) < 0</p>
<p>81 - 4c < 0</p>
<p>c > 81/4</p>
<p>Was this on a practice test? I really don’t think the SAT would ask something like this. They tend not to ask math questions that require you to know obscure things because then it’s no longer testing your ability to do the math. There are probably many people (myself included) who could do the math if they understood the question, but don’t know the term “complex conjugate.” I don’t think I saw anything like this on the SAT. Did you, rspence?</p>
<p>Not on the SAT, it doesn’t test anything on complex numbers.</p>
<p>Btw, a “complex conjugate” of z = a + bi is defined as \bar{z} = a - bi (a bar over the z), so that z times its conjugate is a real number.</p>
<p>It was on the SAT 2.</p>
<p>But wait, why is the determinant less than zero? Is it because the two non-real roots are always less than 0?</p>
<p>Ok. So think about this geometrically. When will you have real roots? When the graph of y = ax^2+bx+c intercepts the x-axis 2/1 time(s). This occurs if and only if the determinant is greater than or equal to zero. Use r(e^itheta) for a more concrete explanation.</p>
<p>Whoops, “discriminant.” “Determinant” is for square matrices.</p>
<p>Recall the quadratic formula, for ax^2 + bx + c = 0, x = (-b ± sqrt(b^2 - 4ac))/2a. You have two non-real roots if and only if b^2 - 4ac < 0.</p>
<p>I just realized, the discriminant “could” equal zero, because the complex conjugate of a real number, say 7, is itself. So c could equal 81/4. Small error in the question.</p>