Higher Maths 2012

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woke up bright eyed and bushy tailed, got a good feeling today this is going to be a nice paper! they can't put us through the torture from that credit exam from last year and a hard exam this year?! haha good luck everyone!

Reply 681

lozza.22

Good luck today everyone! Bit nervous but sure everyone will do fine!!

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Reply 682

Blue7195

just as short revision, when

b^2-4ac > 0

then the roots are real, irrational if say

\sqrt{23}

, and rational if

\sqrt{25}

? then, if

b^2-4ac = 0

then the roots are equal, and if

b^2-4ac <0

then the roots are not real, unequal, ah good remembered !

Reply 683

member592149

Up at 8 for a couple past papers and reminding myself of some of my notes. :colone:

Reply 684

Revolver72

Best of luck today, all :biggrin:

Reply 685

rawragee

I tried to do a supposedly nice past paper to get me in a mathematical mindset I guess, then started crying because I couldn't do the first steps of an 8 marker.. awh that's a good omen! :redface:

Good luck today everybody, we've all worked hard enough! :biggrin:

Reply 686

TheRedGoldfish

Original post by Blue7195

just as short revision, when

b^2-4ac > 0

then the roots are real, irrational if say

\sqrt{23}

, and rational if

\sqrt{25}

? then, if

b^2-4ac = 0

then the roots are equal, and if

b^2-4ac <0

then the roots are not real, unequal, ah good remembered !

Remember that when working out complicated roots you are square rooting b^2 - 4ac

In 2011 at least, the discriminant = 23, in other words b^2 - 4ac = 23 and since it's greater than 0 then there will be two real roots

When it equals 0 there is one equal root and if it's less then there will be no real roots

And if a surd still remains even after simplifying then yes it's irrational and if it can eventually be broken down into whole numbers then it is rational

(edited 11 years ago)

Reply 687

Blue7195

okay please cn someone help me?!

with this? ahhhh

You need to put the two equations equal to each other and solve to find x.

Reply 690

auberjean

Original post by Blue7195

with this? ahhhh

Start by setting the equations equal to each other and solve to find x.

Reply 691

Blue7195

Original post by auberjean

Start by setting the equations equal to each other and solve to find x.

i end up getting

x^{3} + 6x^{2} -37x +30

.. how do i solve this? synthetic doesn't work.. neither does diff..

Reply 692

aroy45

Original post by Blue7195

i end up getting

x^{3} + 6x^{2} -37x +30

.. how do i solve this? synthetic doesn't work.. neither does diff..

Make x=1 and do synthetic division

Reply 693

Herzschmerz

EDIT: The person above is right.

(edited 11 years ago)

Reply 694

Blue7195

Original post by aroy45

Make x=1 and do synthetic division

thanks so much! seen it and was like ahhh... incase it came up today so freaked out a lil' :P

Reply 695

DarrenBoyle

Original post by aroy45

Make x=1 and do synthetic division

Yep, 1 is a root and thus (x-1) is a factor. 3 is also a root.

Reply 696

tomctutor

Original post by Blue7195

i end up getting

x^{3} + 6x^{2} -37x +30

.. how do i solve this? synthetic doesn't work.. neither does diff..

x =0: y =30 no good
x =1: y =1+6-37+30 = 0 is good
(x-1) a factor find the others through synthetic division:

1|_1_6_-37_30
_|___1___7_-30
___1_7_-30_|0

so now factorise