The problem has a large square root which you can't type here so I'm just posting a picture of what the equation looks like:
http://i24.photobucket.com/albums/c17/sf2icons/Untitled-2.png
Would you use completing the square? I tried to factor but none of them would.
2007-05-21 14:35:48 · 3 answers · asked by Sabrina F 2 in Science & Mathematics ➔ Mathematics
My first instinct is to square both sides. This should get rid of the square root.
cos(60) squared is 0.25 (if 60 is in degrees)
None of the quadratics factor in real numbers (they would in complex numbers, but I've tried and it is not helpful).
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You'll end up with finding the roots in a 4th degree polynomial:
1/4 = (x^2 -8x +40)^2 / (x^2 -16x +80)(x^2 + 100)
(x^2 -16x +80)(x^2 + 100) = 4(x^2 -8x +40)^2
x^4 -16x^3 +180x^2 -1600x + 8000 = 4x^4 -64x^3 + 576x^2 -2560x +6400
0 = 3x^4 -48x^3 +396x^2 -960x -1600
(That is: if I didn't get mixed up in adding...)
2007-05-21 14:45:00 · answer #1 · answered by Raymond 7 · 0⤊ 0⤋
yea i would say try completing the square. but i'm also thinking that the cos(60) is equal to 1/2 b/c cos of 60 degrees is the same as saying the cosine of pi/3.
2007-05-21 14:44:58 · answer #2 · answered by lion_dancer_boi062 2 · 0⤊ 0⤋
Factor all of them, then just multiply and find x.
EDIT: sorry i didn't see that you said they wouldn't factor. if they don't factor then use the quadratic formula.
2007-05-21 14:45:58 · answer #3 · answered by Heather 6 · 0⤊ 0⤋