solve the following question

Given that (5 − x)2 = y − 20 2 where x and y are positive integers, find the value of
x and the value of y.

Given that (5 − rootx)squared = y − 20root2 where x and y are positive integers, find the value of
x and the value of y.
Sorry not working from my laptop and cant find root or squared button.

A similar question to this one actually came up on my IGCSE Additional Math exams a few weeks ago. Thankfully I was learning ahead and was rather familiar with complex numbers and managed to do the analogue in this situation. That is, equating the 'real' and 'imaginary' parts, where real is just the non-surd part and imaginary is the surd part of each side of the equation. Interesting question though! Is the answer then:

?

One thing I don't understand is how you can be sure that this is a unique solution? In my exam, I got a quadratic and hence got two pairs of answers. But how can you ensure that they are the only solution(s)?

my reasoning with this was that:

expanding the lhs and re-arranging the equation, you have that:

$y=(25+x)+10(\sqrt{8}- \sqrt{x})$

clearly, the $25+x$ is an integer iff x is, so that only leaves that you HAVE TO HAVE:

$\sqrt{8}- \sqrt{x}$ is a positive (i.e. non negative i.e. including zero) integer, which is only possibly if:

$\sqrt{8}- \sqrt{x}=0$ since $\sqrt{2}$ is not an integer even if $x$ is a square.

(p.s haven`t looked at the links)

Ooh, that's actually quite interesting. One thing I can't quite grasp about your argument is why there needs to be a restriction on why $\sqrt{8} - \sqrt{x}$ needs to be an integer.

My reasoning was that $y$ would remain an integer if $\sqrt{8} - \sqrt{x} = \frac{k}{10} \forall k \in \mathbb{N}$, this would lead to $k = 20\sqrt{2} - 10\sqrt{x}$ and, continuing your argument, we would set $20\sqrt{2} - 10\sqrt{x} = 0 \Rightarrow x=8$

Is that correct?

that`s a good way of putting it! more "mathematically rigorous" than mine (but for "some" k in N, `cause, to start with you`re only looking for one instance), i guess! - then you`d follow with the restriction that this is true if and only if k=0.

(i just noted that 10 is an integer, so what`s inside the bracket has to be)

A similar question to this one actually came up on my IGCSE Additional Math exams a few weeks ago. Thankfully I was learning ahead and was rather familiar with complex numbers and managed to do the analogue in this situation. That is, equating the 'real' and 'imaginary' parts, where real is just the non-surd part and imaginary is the surd part of each side of the equation. Interesting question though! Is the answer then:

?

One thing I don't understand is how you can be sure that this is a unique solution? In my exam, I got a quadratic and hence got two pairs of answers. But how can you ensure that they are the only solution(s)?

Hi Brooke,
Recheck the question again. Do u mean (5 - 2root x) or exactly wat u wrote
Thanks

Hi Can you write out the method you used to find the answer. Thanks

Hi

Thanks

I am learning maths so why does
$-10\sqrt{x} = -20\sqrt{2} \Rightarrow \sqrt{x} = 2\sqrt{2}$

Oh snap! You didn't do the question yet? Gah, I shouldn't have posted the solution. - I thought you'd already done the question.
Anyways,


$-10\sqrt{x} = -20\sqrt{2}$, divide both sides by $-10$, we get:

$\frac{-10\sqrt{x}}{-10} = \frac{-20\sqrt{2}}{-10}$, and since $\frac{-20}{-10}= 2$, the above equation turns into: $\sqrt{x}=2\sqrt{2}$