# Chords easy question --Test tomorrow

• May 19th 2010, 08:27 PM
Chords 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, 08:48 PM
pickslides
Quote:

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.

Read this Intersecting Chord Theorem - Math Open Reference

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

Solve for $x$.
• May 19th 2010, 08:57 PM
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, 09:01 PM
pickslides
Quote:

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

What is happening between these steps?

Quote:

0 = x^2 + x - 10
0 = (x -10) (x + 1)

Sorry my friend, this is all types of wrong. You really should revise some factorisation.
• May 19th 2010, 09:29 PM
$0 = 2x^2 + x - 10 \implies x =\frac{-1\pm\sqrt{1^2-4\times 2 \times -10}}{2\times 2}=\dots$