Help with FSMQ Polynomial question?

Hi, I have been given this question:



I have arrived at an answer, however, I do not know if it was the correct one. Here are my workings:

1) Using Pythagoras' Theorem, I found the longest side of the triangle x and (x + 8) would make, giving me the root of x squared plus (x + 8) squared.
2) Then, I used Pythagoras' Theorem again with (x + 1), giving me the root of (previous answer) squared plus (x + 1) squared.
3) All of this equals thirteen.
4) I then squared both sides, removing the first root and making it equal to 196
5) Because the answer to (1) was a root squared, it was equal to what is inside the bracket.
6) I then multiplied out the (x + 8) squared and the (x + 1) squared.
7) Collecting like terms, this gave me (3x squared + 18x - 104) = 0
8) Using the quadratic formula, I eventually got the answers 3.6 and -9.6 (1.d.p)

Is this the correct way of doing it? If not, could you please give me one or two hints?

Thanks,

Hariex

Do you know about the 3D version of Pythagoras? Your answers don't look quite right to me.

This link should help

Was this covered in GCSE?

I covered it, but it was very brief and only about finding the length of a diagonal given the sides of the box.

I was taught that (using the Wolfram demonstration) you would use 2D Pythagoras, using x and y to find a, then you use the Pythagoras again, using a and z to find d.

Is this incorrect? (And do you know what I'm on about? )

You can just use $d^2 = a^2 + b^2 + c^2$ where d is the space diagonal (the pole).

Thanks.

Is this the same as my method in the previous post? Will they give different results?

I haven't checked your work but it is possible to use 2D Pythagoras twice. This is all covered in the FSMQ textbook - do you have a copy?

Yes I do have the textbook - I'll look over it.