Trial & Improvement

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June 10th 2007, 03:59 AM
Danielisew

Trial & Improvement

I don't know where to start :D
I started at 1.5, and i think i was doing the calculations wrong..

You can use a calculator for this:

Q1: The equation x^3 + 10x = 21
has a solution between 1 and 2

Use a trial and improvement method to find this solution.
Give your answer correct to one decimal place
You must show ALL your working.

Can someone do this for me please, and make it clear what you type into your calculator paper?

Also i don't get why this is worth 4 marks:

Q2: Ann Drives 210 km in 2 hours 40 minutes.

Work out Ann's average speed.

Isn't it just 210.2.4 ?
Because the ratio of km to hours is the same as metres to seconds

ps. I got my GCSE maths calc paper tomorrow, the non calc went very well - thanks for your help on that! :)
June 10th 2007, 05:02 AM
topsquark

Quote:

Originally Posted by Danielisew

I don't know where to start :D
I started at 1.5, and i think i was doing the calculations wrong..

You can use a calculator for this:

Q1: The equation x^3 + 10x = 21
has a solution between 1 and 2

Use a trial and improvement method to find this solution.
Give your answer correct to one decimal place
You must show ALL your working.

Start with x = 1.5 ==> $1.5^3 + 10 \cdot 1.5 = 18.375$

So you need to find a larger number. Now split (roughly) the interval from 1.5 to 2: use x = 1.7 (or 1.8 or even 1.75 if you wish)
x = 1.7 ==> $1.7^3 + 10 \cdot 1.7 = 21.913$

So you need to find a smaller number. Now split the interval from 1.5 to 1.7: use x = 1.6
x = 1.6 ==> $1.6^3 + 10 \cdot 1.6 = 20.096$

etc.
I get that x is between 1.6503 and 1.6505, so either way we have x is about 1.6 (or 1.7 if you round it properly.)

-Dan
June 10th 2007, 05:06 AM
topsquark

Quote:

Originally Posted by Danielisew

Q2: Ann Drives 210 km in 2 hours 40 minutes.

Work out Ann's average speed.

Isn't it just 210.2.4 ?
Because the ratio of km to hours is the same as metres to seconds

$1 \, km/hr = 0.27778 \, m/s$
Where did you hear that they are in the same ratio??

So Ann drives 210 km = 210000 m in 2 hr + 40 min = 160 min = 9600 s.

Thus Ann's average speed was:
$v = \frac{210000 \, m}{9600 \, s} = 21.875 \, m/s$

-Dan
June 10th 2007, 05:27 AM
Danielisew

looking at the question do you need to do in m/s?
I got 87.5 km/h which sounds about right..
June 10th 2007, 05:34 AM
topsquark

Quote:

Originally Posted by Danielisew

looking at the question do you need to do in m/s?
I got 87.5 km/h which sounds about right..

Careful! Your time period is 2 hr and 40 min. This is NOT 2.4 hr as you seem to be claiming.

40 min = 40/60 hr = 0.666667 hr.

So your average speed in km/hr will be
$\frac{210 \, km}{2.666667 \, hr} = 78.75 \, km/hr$

-Dan
June 10th 2007, 05:42 AM
Danielisew

ok Thanks.. moving onto my new question i'm stuck on now:

8. Lisa said that - 2 is the only value of x that satisfies the equation x^2 + 4x + 4 = 0

Was Lisa correct?
Show your working to justify your answer.

I just don't know where to start on these kind of questions... :(

Help please? :)
June 10th 2007, 05:59 AM
Danielisew

aarrggh this paper is proving to be really hard! Got another question I'm stuck on...

9. Bytes is a shop that sells computers and digital cameras

In 2003, Bytes sold 620 computers.
In 2004, Bytes sold 708 computers.

a) Work out the percentage increase in the number of computers sold.
Give your answer to an appropriate degree of accuracy.
June 10th 2007, 06:36 AM
Danielisew

I got question 9, still stuck on one before that!
June 10th 2007, 07:53 AM
topsquark

Before I do anything else, please let me say that you should start a new thread when asking new questions.

-Dan
June 10th 2007, 07:57 AM
topsquark

Quote:

Originally Posted by Danielisew

ok Thanks.. moving onto my new question i'm stuck on now:

8. Lisa said that - 2 is the only value of x that satisfies the equation x^2 + 4x + 4 = 0

Was Lisa correct?
Show your working to justify your answer.

I just don't know where to start on these kind of questions... :(

Help please? :)

There are two ways of looking at this.

First, if you know that x = -2 is a solution to the polynomial equation (which it is) then you can state that x - (-2) = x + 2 is a factor of the polynomial $x^2 + 4x + 4$ . So you can divide (either long division or synthetic division) to find that
$x^2 + 4x + 4 = (x + 2)(x + 2)$

So x + 2 = 0 and x + 2 = 0 give us our solutions. In both cases x = -2.

Second, you can simply solve the equation:
$x^2 + 4x + 4 = 0$
by your favorite method of solving quadratics. You will get that x = -2 is a root of multiplicity 2.

Does this mean there are two solutions? It depends on who you talk to. (Some would say repeated roots are really separate solutions to the equation.)

Either way there we can positively state that there is no number other than x = -2 that solves the equation.

-Dan
June 10th 2007, 08:17 AM
Danielisew

THanks :)

I am called Dan! :(
June 10th 2007, 08:32 AM
topsquark

Quote:

Originally Posted by Danielisew

THanks :)

I am called Dan! :(

It's a good name! :)

-Dan
June 10th 2007, 09:11 AM
CaptainBlack

Quote:

Originally Posted by Danielisew

Q1: The equation x^3 + 10x = 21
has a solution between 1 and 2

Use a trial and improvement method to find this solution.
Give your answer correct to one decimal place
You must show ALL your working.

We seek a root of $f(x)=x^3+10x-21$ between 1 and 2:

Using the bisection algorithm we get:

Code:

lo f(lo) hi f(hi) mid f(mid) 1 -10 2 7 1.5 -2.625 1.5 -2.625 2 7 1.75 1.859 1.5 -2.625 1.75 1.859 1.625 -0.4589 1.625 -0.4589 1.75 1.859 1.688 0.6804 1.625 -0.4589 1.688 0.680 1.641 -0.1755 1.641 -0.1755 1.688 0.680 1.649 -0.0306 1.649 -0.0306 1.688 0.680 1.653 0.0421 1.649 -0.0306 1.653 0.0421 1.651 0.0103 1.649 -0.0306 1.651 0.0103 1.650 -0.0079 1.650 -0.0079 1.651 0.0103

So to one decimal place the required root is 1.7 (that is the root
is bigger than 1.650 and less than 1.651, all of which round to 1.7

RonL