# Just need help with these 2 equations

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• Mar 11th 2007, 05:28 PM
Just need help with these 2 equations
1.Find all solutions of the equation http://hosted.webwork.rochester.edu/...146179e861.pngand express them in the form http://hosted.webwork.rochester.edu/...b3618a6031.png
solutions:

2.Two ships leave a harbor at the same time, traveling on courses that have an angle of 130degrees between them. If the first ship travels at 22 miles per hour and the second ship travels at 28 miles per hour, how far apart are the two ships after 2.9 hours?
• Mar 11th 2007, 05:49 PM
Jhevon
Quote:

1.Find all solutions of the equation http://hosted.webwork.rochester.edu/...146179e861.pngand express them in the form http://hosted.webwork.rochester.edu/...b3618a6031.png
solutions:

2.Two ships leave a harbor at the same time, traveling on courses that have an angle of 130degrees between them. If the first ship travels at 22 miles per hour and the second ship travels at 28 miles per hour, how far apart are the two ships after 2.9 hours?

1. x^2 + 4x + 5 = 0
=> x = [-4 +/- sqrt(16 - 4(5))]/2 ......by the quadratic formula
=> x = [-4 +/- sqrt(-4)]/2
=> x = (-4 +/- 2i)/2
=> x = -2 + i or -2 - i

2. I will attach a figure soon. notice that if the first ship travels at 22 mph for 2.9 hrs, it will cover a distance of 63.8 miles. if the second ship travels at 28 mph for 2.9 hours, it will cover a distance of 81.2 miles. so we have a triange with two adjacent sides, one 63.8 miles in length, the other 81.2 miles in length and an angle of 130 degrees between them. we want to find the length of the side opposite to the 130 degree angle. we use the cosine rule (or the law of cosines as some call it).

by the cosine rule:
a^2 = b^2 + c^2 - 2bc*cos(A)
=> a^2 = (63.8)^2 + (81.2)^2 - 2(63.8)(81.2)cos(130)
=> a^2 = 10663.88 - 10361.12*cos(130)
=> a^2 = 10663.88 - 10361(-0.64279)
=> a^2 = 17323.904
=> a = sqrt(17323.904)
=> a = 131.62 miles

therefore the distance between the ships after 2.9 hrs is 131.62 miles
• Mar 11th 2007, 05:53 PM
Jhevon
Quote:

Originally Posted by Jhevon
1. x^2 + 4x + 5 = 0
=> x = [-4 +/- sqrt(16 - 4(5))]/2 ......by the quadratic formula
=> x = [-4 +/- sqrt(-4)]/2
=> x = (-4 +/- 2i)/2
=> x = -2 + i or -2 - i

2. I will attach a figure soon. notice that if the first ship travels at 22 mph for 2.9 hrs, it will cover a distance of 63.8 miles. if the second ship travels at 28 mph for 2.9 hours, it will cover a distance of 81.2 miles. so we have a triange with two adjacent sides, one 63.8 miles in length, the other 81.2 miles in length and an angle of 130 degrees between them. we want to find the length of the side opposite to the 130 degree angle. we use the cosine rule (or the law of cosines as some call it).

by the cosine rule:
a^2 = b^2 + c^2 - 2bc*cos(A)
=> a^2 = (63.8)^2 + (81.2)^2 - 2(63.8)(81.2)cos(130)
=> a^2 = 10663.88 - 10361.12*cos(130)
=> a^2 = 10663.88 - 10361(-0.64279)
=> a^2 = 17323.904
=> a = sqrt(17323.904)
=> a = 131.62 miles

therefore the distance between the ships after 2.9 hrs is 131.62 miles

here's the diagram, i labeled the points a,b,c and the angles A,B,C
• Mar 11th 2007, 05:53 PM
thanks sooo much. Still it says the first one is wrong? Are you sure that's right?
• Mar 11th 2007, 05:59 PM
Jhevon
Quote:

thanks sooo much. Still it says the first one is wrong? Are you sure that's right?

yup. pretty sure that's right. you can check for yourself by using the quadratic formula. but let's try another method to be sure. how about completing the square.

x^2 + 4x + 5 = 0
=> x^2 + 4x = -5
=> x^2 + 4x + (2)^2 = -5 + (2)^2
=> (x + 2)^2 = -1
=> x + 2 = +/- sqrt(-1)
=> x = -2 +/- sqrt(-1)
=> x = -2 +/- i

which is what we got doing it by the quadratic formula
• Mar 11th 2007, 06:27 PM
qbkr21
• Mar 11th 2007, 06:43 PM
ThePerfectHacker
Quote:
Can it solve cubics?
Try it (I do not have a TI because I hate calculators, hate them with a love).

Furthermore, let me warn you more complicated equations are solved through Newton's method.

They fall as quicky as did Rome if you are not careful.
• Mar 11th 2007, 06:44 PM
Jhevon
Quote:
what program/utility is this? o it's a TI calc.
• Mar 11th 2007, 06:49 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
what program/utility is this? o it's a TI calc.

It is a TI calculator.

I know because it makes my blood run cold as did the "Vulture Eye".

(Or qbkr took a digital photo, gave it to the computer, then posted it here. But I do not think qbkr is that nice to do something like that).
• Mar 11th 2007, 06:56 PM
qbkr21
Re:
This is the TI Voyage 200 Voyage™ 200 by Texas Instruments - US and CANADA . Massive screen with lightning speed. I guess it can solve cubics. I am showing you two screens because the result is so long. Also I just connected my calc to my computer through the USB it came with.

http://item.slide.com/r/1/122/i/gFSx...qjvJSXIQvB2Kg/

http://item.slide.com/r/1/1/i/2ngFOn...W9Kx-MbA5Vq1W/

They do not sell these calculators in stores (Best Buy, Walmart, ect.) You must buy them through an authorized dealer. I drove an hour and a half to pick this one up before my Combinatorics exam.
• Mar 11th 2007, 06:59 PM
ThePerfectHacker
Tell it to solve,
x^3+x=1

I want to see how it presents the answer.
(Because I have done this by hand and It leads to a nested radical).
• Mar 11th 2007, 07:02 PM
qbkr21
Re:
Hey ThePerfectHacker,

In looking over your signature I just wanted to know how Ken was?:D
• Mar 11th 2007, 07:12 PM
qbkr21
Re:
This probably isn't what you are looking for. The square come out as an irrational fraction, but for this particular problem I am getting a complex #.

http://item.slide.com/r/1/69/i/vnUnr...knKM_LTRvQh6S/

http://item.slide.com/r/1/165/i/AesD...O-a2irOLeKU6G/
• Mar 11th 2007, 07:22 PM
ThePerfectHacker
Quote:

Originally Posted by qbkr21
This probably isn't what you are looking for. The square come out as an irrational fraction, but for this particular problem I am getting a complex #.

It seems it is not programmed to solve by radicals.
• Mar 11th 2007, 08:08 PM
qbkr21
Re:
How do you solve cubics what is the formula?

x^2+x=1
x^2+x-1=0

Then what's the next step?

:confused:
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