Series (1 Viewer)

[crammy90]

ANSWERS for b
i) 224
ii) 4392

they use a = 20 and d=12
so can a be different?
cuz then wouldnt that change the t2/t1 = t4/t3 :S

another Q
"show that equation x^2 + (p-3)x - (2p+1) = 0, where p is real, has real distinct roots"
delta = (p-3)^2 - (4*1*-[2p+1])
= (p-3)^2 - (4[-2p-1])
= (p-3)^2 + 8p + 4
= p^2 - 6p +9 + 8p + 4
= P^2 + 2p + 13
+1 = p^2 + 2p + 1 (completing the square???) + 13
= (p-1)^2 + 12
EDITTED:
therefore at its minimum delta = 12 (i.e. 0^2) so it is always greater than 0
therefore real and different roots
where do i go wrong :S

 

[Js^-1]

Hey I got that
p² + 2p +13 > 0
p² + 2p +1 +12 > 0
(p+1)² + 12 > 0
(p+1)² > -12
This is where i get a bit messed up because p is real...

 

[Azreil]

Okay. In the first question, in any series, a can be a different number to d.

In the second question, The working of Js^-1 is right. Until you move 12 across.

Delta tells us the nature of the roots.

If delta < 0, there are no real roots. If delta = 0, there are real roots, but they are the same -- ie, it touches rather than cuts the x axis. If delta > 0, it has distinct, real roots -- ie, it cuts the x axis in two places.

In this case, the smallest number anything squared can be is 0. So delta, at it's smallest, is (-1 + 1)^2 +12 -- ie, at it's smallest it is positive. It therefore has real, distinct roots.