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Other Topics => Completely Off-Topic => Topic started by: whitehouses81680 on December 29, 2005, 03:04:17 pm
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I NEED HELP!
does anyone know how to do this math problem:
22. Find all the values of k such that the expression 3x^2-2x+k=0 has imaginary roots.
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I SHOULD HAVE POSTED THIS ON THE OTHER TOPIC...SRY
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if i am thinking correctly you should use the square root of b squared minus (4)(a)(c).
which would give you all numbers k> -1 would be the answer....
EDIT:
uh, I didn't finish working out the problem... and Will actually finished. (obviously, I am not going to be a math major). So, look at Will's.
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3*x^2 - 2*x + k = 0
f(x) = 3*x^2 - 2*x + k
As you recall, there exists a formula to find the roots of all quadratic equations in the form of a*x^2 + b*x + c = 0. The formula is,
x = (-b (+/-)(d)^(1/2)) / (2*a)
where d = b^2 - 4*a*c.
The only place in this formula that would give rise to an imaginary number (and hence, give us a solution with a non-zero imaginary component) is the radical, if and only if the quantity under it is negative. So, f(x) has roots with a non-zero imaginary component where d < 0 (d is negative). We are already given a and b. We'll plug those into the formula for d and substitute that into the inequality I stated above. Then, we'll solve for c, which is k in the original statement of the problem.
a = 3
b = -2
c = k
b^2 - 4*a*c < 0
4 - 4*3*k < 0
-12*k < -4
k > (1/3)
Therefore, f(x) has roots with a non-zero imaginary part where k > (1/3).
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wouldn't that be 1/3?
where did you get 1/4?
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Made a mistake. Fixed.
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lol well, at least you finished working it out...i feel rather dumb...
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WILL, LIMETWISTER, YOU ALL ROCK...thanks a bunch<3
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and just to throw it out there if anyone is really bored:
The accompanying diagram shows a circular machine part that has rods PT and PAR attached at points T, A, and R which are located on teh circe: mTA:mAR:mRT-1:3:5; RA=12 centimeteres; and PA=5 centimeters
Find the measure of angle P in degrees and the find the length of rod PT to the nearest tenth of a centimeter....
i got that P= 80 degrees
i dont know how to find PT?
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Ok... I was gonna do the problem, but I have a question. You said "mTA:mAR:mRT-1:3:5." I read that as a ratio of the three lengths (1:3:5). However, if I do that, mRT = 20 cm, mRA = 12 cm, and mTA = 4 cm. 12+4 = 16 < 20, so that isn't a triangle. Could you explain what you mean?
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mTA:mAR:mRT-1:3:5<--- thats a circle. The whole circle is broken up...
idk how i can get the diagram on the site :(
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Scan it.
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mTA:mAR:mRT-1:3:5<--- thats a circle. The whole circle is broken up...
idk how i can get the diagram on the site :(
So, mTA and the like denote arcs? The ratio of their lengths is 1:3:5?
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the answer is 9.2 sam, but i don't know why
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what math did u show?