c1 edexcel - another series question :@

Question:

On Alice's 11th birthday she started to recieve an annual allowance. The first annual allowance was £500 and on each following birthday the allowance was increased by £200.

a) Show that, immediately after her 12th birthday, the total allownace that Alice had recieved was £1200.

easy.. a+a+d = 500+500+200 = 1200 squids

b) Find the amount of Alice's annual allowance on her 18th birthday.
easy... a + (n-1)d => 500 + 7 x 200 => 1900 squids

c) Find the total of the allowances that Alice had recieved up to and including her 18th birthday.

easy... n = 8, a = 500, d = 200
Sn = 1/2 n[2a + d(n-1)]
S8 = 4[2x500+200x7] => S8 = 4[1000 + 1400] => 4 x 2400 = 9600 squids

d) When the total of the allowances that Alice had recieved reached £32 000, the allowance stopped.
Find how old Alice was when she recieved her last allwance.
not so easy.. 32000 = 1/2n[1000 + 200n - 200] => 32000 = 1/2n [800 + 200n]
32000 = 400n + 100n^2
100n^2 + 400n - 32000
n^2 + 4n - 320 = 0
(n + 2)^2 = 324
n + 2 = +- sqrt(324)
n = -2 +- 18
n = -2 + 18 = 16
n = -2 - 18 = 20

there are two values for n, which one do i choose and why? thx

(btw anyone know a way to work out square roots of 3 digit numbers without using a calculator?)

ah yeah ur both right :P its actually -20 not 20 (silly mistake ).. and i think u can only leave answers in surd form if they arent solvable without a calculator..?

(btw anyone know a way to work out square roots of 3 digit numbers without using a calculator?)

OK, my method is not exactly scientific, but I use it to simplify surds and find square roots...

Do you know how to express a number as a product of prime factors? I'm guessing you do. Sometimes you don't need to go all the way, let me show you using your example of sqrt(324).

324 / 2 = 162
162 / 2 = 81

I stopped here becuase everybody knows the square root of 81.

So know you know that $324 = 2^2.81$

Therefore you know that $\sqrt(324) = 2.\sqrt(81)$

and therefore the square root of 324 is 18.

The same principle can be used for larger numbers, take for example 104976.

Express it as a product of prime factors, in this case it comes to,

$104976 = 2^4.3^4.9^2$

So, to take the square root of 104976 you would "half" the indices, giving,

$\sqrt(104976) = 2^2.3^2.9 = 324$


EDIT: Beaten to it.

I have a question.

Why can't I factorise it like this.

n(n-4) = 320

Whats wrong with the maths here?

why cant you factorise it like that lol? that looks fine to me

But the answer is 324 or 320. Which isn't quite the right answer.

Hence the dilema.

lol the expression isnt equal to 0, so you cant say one of the brackets has to equal the answer.. if you multiply two numbers together, at least one of them has to be equal to 0 for the expression to equal to 0 overall.. i guess the difference is, 320 has many different factors, where as with 0, the one of the factors is always 0..

i get two values for n.. i know i add n to 10 once i get it, right?