Practice Test Issues...
There are a few problems on the practice test that are giving me a hard time:
6. This problem seems like it should be pretty easy, but I'm stuck on it. I tried several different methods, and everytime I tried to solve for b on my calculator, it would say "true" or "false" instead of giving me a value. In my most recent attempt, I defined f(x)=x^3+b*x^2-3*x+d on my calc, and then took f(2+5i) and set the answer [-21b+d-148+(20b-80*i] aside. I then took f(-4) and solved it for d. Since f(-4)=16b+d-52 and f(-4) also equals 0, d equals 52/(16b). I then tried plugging this value in for d in the equation -21b+d-148+(20b-80)*i=0 and solving for b, but my calculator simply replied, "false."
11a. and 11b. This question is daunting, although I should probably give it a second try. I have a feeling it relates to that grid/vector discussion we had in class the other day, and that whole thing went right over my head. If anyone understand this I could REALLY use your help!
13. Like Carrie, I honestly don't know where to begin.
ANY help would be GREATLY appreciated!!
6. This problem seems like it should be pretty easy, but I'm stuck on it. I tried several different methods, and everytime I tried to solve for b on my calculator, it would say "true" or "false" instead of giving me a value. In my most recent attempt, I defined f(x)=x^3+b*x^2-3*x+d on my calc, and then took f(2+5i) and set the answer [-21b+d-148+(20b-80*i] aside. I then took f(-4) and solved it for d. Since f(-4)=16b+d-52 and f(-4) also equals 0, d equals 52/(16b). I then tried plugging this value in for d in the equation -21b+d-148+(20b-80)*i=0 and solving for b, but my calculator simply replied, "false."
11a. and 11b. This question is daunting, although I should probably give it a second try. I have a feeling it relates to that grid/vector discussion we had in class the other day, and that whole thing went right over my head. If anyone understand this I could REALLY use your help!
13. Like Carrie, I honestly don't know where to begin.
ANY help would be GREATLY appreciated!!
posted by Casey Blue at 2:44 PM
1 Comments:
For number 6 I was stuck at first too. What i did was i took the fact that if there is a complex root in order to have real coefficients the conjugate must also be a root. therefore, I expanded (x+4)*(x-(2+5i))*(x-(2-5i)) and got a coefficient of 0 for the x^2 term.
For 11a I didn't really do a proof but what i did was i said that z=2+3i and w=5+7i. These values are random. i then said that z-w=-3-4i then i said the absolute value of that is 5 (i did this part on my calculator) i then used the distance formula and called z and w coordinates which gave the same end result of 5
For 11b i was very confused at first also. However, the shape question reminded me of the excersize in class when we made all of those regular polygons which made me think that the shape could be a line segment. I first solved for z to get -2(√3 -1) and 2(√3+1). I plugged these values in for z and set the equation equal to zero. This resulted with -2√3-8-2i and 2√3-4-2i. I called these ordered pairs in the complex plane which means that the shape is a line segment with a slope of zero.
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