Normally i wouldn't ask someone to reply quickly, but my test i tomorrow

all help is gratfully appreiciated if you can answer it by midnight

I have made a picture of my problem.

Printable View

- May 19th 2010, 07:27 PMshawnCanadaChords easy question --Test tomorrow
Normally i wouldn't ask someone to reply quickly, but my test i tomorrow

all help is gratfully appreiciated if you can answer it by midnight

I have made a picture of my problem. - May 19th 2010, 07:48 PMpickslides
Read this Intersecting Chord Theorem - Math Open Reference

Therefore in question 2: $\displaystyle 2\times 5 = x(2x+1)$

Solve for $\displaystyle x$. - May 19th 2010, 07:57 PMshawnCanada
Sorry i did not have tiime to explain my attempt:

5x2 =x(2x+1)

10 = 2x^2 + x

0 = 2x^2 + x - 10

2 / 0 = 2 / (x^2 + x - 10)

0 = x^2 + x - 10

0 = (x -10) (x + 1)

(split into two equation)

x = 10 <---

x = -1

In the end im left with x = 10, when in my textbook x = 2 ? - May 19th 2010, 08:01 PMpickslides
- May 19th 2010, 08:29 PMshawnCanada
There's no way you can do the answer for me?

i know i would learn better on my own but

i'v done 23/24 questions and i just need

you question done so i can understand what

i did wrong here - May 19th 2010, 08:59 PMpickslides
$\displaystyle 0 = 2x^2 + x - 10 \implies x =\frac{-1\pm\sqrt{1^2-4\times 2 \times -10}}{2\times 2}=\dots$