# Thread: Chords easy question --Test tomorrow

1. ## 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.

2. Originally Posted by shawnCanada
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 $x$.

3. 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 ?

4. Originally Posted by shawnCanada
0 = 2x^2 + x - 10
2 / 0 = 2 / (x^2 + x - 10)
What is happening between these steps?

Originally Posted by shawnCanada
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.

5. 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

6. $0 = 2x^2 + x - 10 \implies x =\frac{-1\pm\sqrt{1^2-4\times 2 \times -10}}{2\times 2}=\dots$