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Alan

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 the roots of 3x^2 -2x-5=0 are alpha and beta find the equation whose roots are alpha^2 and beta^2 (Posted for sacha)

So alpha = 5/3 and beta = -1
To get the (quadratic) equation whose roots are alpha2 and beta2 we construct:
(x - alpha2)*(x-beta2) = 0 or:
x2 -(alpha2+beta2)*x + alpha2*beta2 = 0 lire plus ..
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Alan  22 avr. 2014
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Bertie:

...

A man receives a cheque and goes to his local bank to cash it.
The cashier at the bank is having an exceedingly bad day and mistakenly mixes up the
dollars and cents, (pounds and pence in England), giving the man y dollars and x cents
rather than x dollars and y cents. The man doesn't notice the mistake and returns home,
on the way spending 5 cents of the money. (He had no other money with him when he left home.)
Only when he gets home does he check, and he finds that he now has exactly double
the amount on the cheque.
The question is how much was the cheque for ?



Here's my reasoning.

The cheque is worth 100x + y cents, where y must be an integer between 1 and 99 inclusive.
The man is given 100y + x cents, where x is clearly also an integer between1 and 99 inclusive.
When he gets home the man has 100y + x - 5 cents and we are told that this is twice the original value of the cheque, so we must have 100y + x - 5 = 2*(100x + y).

Collecting terms, we can write this as 98y - 5 = 199x ...(1)

Clearly, y must be bigger than x or the two sides will never match!
Suppose y is twice the size of x. 98*(2x) - 5 is 196x - 5 which is smaller than 199x, so y must be more than twice the size of x.

Let y = 2x + n ...(2) where n is another integer.

Put (2) into (1) to get 98*(2x+n) - 5 = 199x or, 196x + 98n - 5 = 199x. This simplifies to: 98n - 5 = 3x ...(3)

Now x must be bigger than n.
Let's try x = 31n in (3) (trying to get close to, but a little below 98n) .
Putting this in (3) we get 98n - 5 = 93n, which simplifies to 5n - 5 = 0.
This gives n = 1, which in turn gives x = 31, and putting these values for n and x into (2) we have y = 2*31+1 or y = 63.

(Note: we could have tried x = 32n in (3) to get us closer to 98n, but we would have found that it was too close and there was no positive integer value of n that would satisfy the resulting equation for n.)

So the original cheque is worth 100*31+63 cents: i.e. 3163 cents or 31 dollars and 63 cents.
5 avr. 2014