Please answer this math question. It's not THAT hard. =)?

Question

If E represents the elevation in meters above sea level and T represents the boiling point of water in degrees Celcius at that elevation, then E and T are related by the equation below.

E = 1000(100-T) + 580(100-T)^2

a. Solve the equation above for T. (Hint: Begin by substituting x for 100-T. Then solve for x.)
b. What is the approximate boiling point of water at an elevation of 1600 meters?

Explanations are very appreciated. Please try and help me with the problem. Thanks!

Answer 1

Ok, first replace (100-T) with x:
580x² + 1000x - E = 0

Now, get the x² term to a factor of 1 by dividing by 580 (reducing the fractions)...
x² + 50x/29 - E/580 = 0

Bring E/580 to the other side and complete the square to find the roots (add 625/841 to both sides)...
x² + 50x/29 + 625/841 = E/580 + 625/841

Now, we have this:
(x + 25/29)² = E/580 + 625/841

Take the square root (plus or minus!)...
x + 25/29 = +/- sqrt(E/580 + 625/841)

Simplify...
x = -25/29 +/- sqrt(E/580 + 625/841)

Replace x with 100-T again, and solve for T...
100 - T = -25/29 +/- sqrt(E/580 + 625/841)
T - 100 = 25/29 +/- sqrt(E/580 + 625/841)

The answer to question a is:
T = 100 + 25/29 +/- sqrt(E/580 + 625/841)


Now for question b...
T = 100 + 25/29 +/- sqrt(1600/580 + 625/841)
T = 102.73°C or 98.99°C

Both of these answers give E = 1600 m (I checked it)... obviously, there cannot be two boiling points, so this equation is not very accurate. However, according to the problem as given, the answer is either 102.73°C or 98.99°C.

Answer 2

E = 1000(100 - T) + 580(100 -T)^2
E = 1000x + 580x^2

Using 1600
580x^2 +1000x - 1600 = 0 divied by 10
This is now a quadraatic.
58x^2 + 100x - 160 = o divide by 2
29 x^2 + 50x - 80 = 0

Use the quadratic equation
x = -b +/- Sq Rt(b^2 - 4ac)
____________________
2a
x = - 50 +/- Sq Rt(50^2 - 4(29)(-80))
___________________________
2 (29)
x = -50 +/- Sq Rt(2500 + 9280)
_______________________
58
x = -50 +/- Sq Rt (11780)
__________________
58
x - -50 +/- 108.5357
_____________
58

x = -50 + 108.5357 & -50 - 1085357
_____________ _____________
58 58
x = 58.5357/58 -158.5357/58
x = 1.009236352 -2.733374283

Hence 100 - T = x
T = 100 - x
T = 100 - 1.009236352
T = 98.99 degrees C.

NB Do NOT use -2.7333 because this would raise the BP of water with increased height and reduced pressure. The boiling point of water decreases with reduuced pressure, which would occur at a height of 1600 metres.

Answer 3

E = 1000(100-T) + 580(100-T)^2
(a) Put (100 -T)=x, then the above Eq.
becomes: E=1000x+580.x^2 or
580.x^2 +1000.x -E =0. This is a quadratic Eq.
so its solution is :
x=(1/1160)(-1000 + or -sq rt(10^6+2320.E)).
Now x=100-T,must be positive
(as T is more than 0 and less than 100),
the negative sign in this expression for x is not acceptable.So the only acceptable solution is
x=(1/1160)(-1000+sq rt(10^6+2320.E))
.which is equal to (100-T). So
T=100 -(1/1160)(-1000+sq rt(10^6+2320.E)).
This is the desired solution.

(b)
Now for E=1600, we calculate
sq rt(10^6+2320.E))=sq rt(10^6 + 2320*1600)=2170.71
So T= 100 -(1/1160)(-1000+2170.71)
=100-1.01=98.99 degrees Celsius.Answer.

Answer 4

let x = 100-T

at sea level, E = 0; then

1000x+580x^2 = 0
58x = -100
x = -100/58

at 1600 m;

1600 = 1000(100-T) + 580 (100-T)^2
using quadratic equation, solve for T and you'll arrive for the answer.

Answer 5

let x = 100-T

at sea level, E = 0; then

1000x+580x^2 = 0
58x = -100
x = -100/58

at 1600 m;

1600 = 1000(100-T) + 580 (100-T)^2
using quadratic equation, solve for T and you'll arrive for the answer.

Answer 6

E = 1000(100-T) + 580(100-T)²

ie 580X² + 1000X - E = 0

So X = [-1000 ±√((1000)² - 4* 580 * (-E))]/(2*580)

= [-1000 ±√(1000000 + 4*580E)]/(2*580)

=[ -500±√(250000 + 580E)]/580

When E = 1600

X = [-500 ±√(250000 + 580*1600)]/580

= [-500 ±√(930500)]/580

≈ 0.80 0R -2.53

So 100 - T = 0.80 0R -2.53

So T = 99.20 or 102.53

Now T = 99.20 when E = +1600 (ie above sea level as air pressure is reduced)
and T = 102.53 when E = - 1600 (ie below sea level as air pressure is increased)

Answer 7

wa...so tedious...expand the equation:

E = 100 000-1000T+580(10 000-200T+T^2)
E = 100 000-1000T+5 800 000-116 000T+580T^2
E = 580T^2+115 000T+5 900 000
E = 4(145T^2+28 750T+1 475 000)
E = 20[ 29T^2 + 5750T + (2875)^2 - (2875)^2+295000 ]
:::(using completing the square method)

E = 20[ (29T + 2875)^2 - 8 265 625 + 295 000]
E = 20 [ (29T + 2875)^2- 7 970 625 ]
E/20 = (29T + 2875)^2 - 7 970 625
::::(divide by 20 on both sides)

E/20 + 7 970 625 = (29T + 2875)^2
::::(add 7 970625 on both sides)

{square root} (E/20 + 7 970 625) = 29T + 2875
{square root} (E/20 + 7 970 625) -2875 = 29T
::::(minus 2875 on both sides)

then divide both sides by 29 to get rid of the 29T to obtain T only.
so,
T = [ {square root} (E/20 + 7 970 625) - 2875] / 29


Therefore, to obtain b), substistute 1600 into T since it is the boiling point of water and manipulate. This is easier than the previous one since it's just substituting into the equation.

Answer 8

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