Am I just being stupid? GCSE Maths help!

ballistic said:
you can do calculate it without a calculator
Click to expand...
yea just remembered I wrote that down, I used two methods: the product,factor and sum one. Wrong

And the quadratics

Hopefully I crossed out the right one :p


People talking about that last question as if it were all together a different kettle of fish made me doubt my methods.


Thanks for making me feel good again :D
 
crunchman said:
I did Higher WJEC so it's probably very different to your exams.

The hardest question for me was:

15x^2-19x-10=0

I can't remember what I wrote down, but I wrote down a lot. My mind was a blank because I totally forgot how to attempt those kind of questions.

I know it's to do with factor, product and sum but I couldn't narrow the answer down as I was using trial and error.

Anyone got a clue ?
Click to expand...

If that was your hardest question I envy you, thats easy compared to the ones on the aqa exam, you just use the quadratic equation, the last few questions we had to do were insane.
 
equation5.gif


depends on the numbers, that's the formula we've been given. I'd rather just do it the simpe way and use that formula for calc papers.

Edit; hotlinked from a math's site. I'm gonna assume they're not going to link it back to goatse ;)
 
I did my Edexcel maths higher paper today (along with geography - but the AQA exam board) and I found it relatively easy. The only question that proved difficult was about 8(Square root 8) along with, and having to work out what K was when it was 8^k. And the making a the subject of the formula I couldn't do, that was too hard. :p But the (n^2 - a ^2) - (n - a)^2 question I managed to nail in the last five minutes I think. :D But my answer as to why the output was an even integer was rushed a bit, but I hope I got it right.
 
crunchman said:
I did Higher WJEC so it's probably very different to your exams.

The hardest question for me was:

15x^2-19x-10=0

I can't remember what I wrote down, but I wrote down a lot. My mind was a blank because I totally forgot how to attempt those kind of questions.

I know it's to do with factor, product and sum but I couldn't narrow the answer down as I was using trial and error.

Anyone got a clue ?
Click to expand...

For a question like this, I'd always starts by checking if it will factorise. Look at b^2 - 4ac. This is 361 + 600, which is 961. This is a perfect square, so it will factorise.

You can continue with the formula, as you've done a lot already, but to use the factorisation method, you're looking for:
(ax + b)(cx + d)

with ac = 15, and bd = -10

Possibilities for a and c are 1, 15 and 3, 5
Possibilities for b and d are 1, 10 and 2, 5 with one of them negative

Now you need to choose the appropriate pairs so that ad + bc = -19

It's not too difficult to see that you need the 3,5 and 2,5 pairs, coupling the two 5s together, and the 2 and 3 together. Moreover, you need 5 x -5, and 2 x 3

This leads to (3x -5)(5x +2) = 0
 
georges said:
equation5.gif
Click to expand...
So you just end up with (19 + 31)/30 and (19-31)/30, which is 5/3 and -2/5. Not too hard without a calculator - hardest bit's probably SQRT(961), but if you made a stab that 30^2 is 900 so it can't be far off, you'd soon find out it was 31^2.
 
Energize said:
If that was your hardest question I envy you, thats easy compared to the ones on the aqa exam, you just use the quadratic equation, the last few questions we had to do were insane.
Click to expand...
That's what I found the hardest.

It was more of a question which pee'd me off because of the multipile ways to answer it.
 
csmager said:
So you just end up with (19 + 31)/30 and (19-31)/30, which is 5/3 and -2/5. Not too hard without a calculator - hardest bit's probably SQRT(961), but if you made a stab that 30^2 is 900 so it can't be far off, you'd soon find out it was 31^2.
Click to expand...

Thing is.. the easier way for simpler stuff is easier ;)

I only use that formula for the more complex ones on calculator papers. You really don't need it at all for non calc papers.
 
Drawoh Tesremos said:
For a question like this, I'd always starts by checking if it will factorise. Look at b^2 - 4ac. This is 361 + 600, which is 961. This is a perfect square, so it will factorise.

You can continue with the formula, as you've done a lot already, but to use the factorisation method, you're looking for:
(ax + b)(cx + d)

with ac = 15, and bd = -10

Possibilities for a and c are 1, 15 and 3, 5
Possibilities for b and d are 1, 10 and 2, 5 with one of them negative

Now you need to choose the appropriate pairs so that ad + bc = -19

It's not too difficult to see that you need the 3,5 and 2,5 pairs, coupling the two 5s together, and the 2 and 3 together. Moreover, you need 5 x -5, and 2 x 3

This leads to (3x -5)(5x +2) = 0
Click to expand...

Cheers for the clear explanation.
 
Lost-Prophet said:
Longer than half an hour loL. I spent the whole 2 hours on in. I didn't get:

Question 23: About 20% wages
24: similar triangles (I think the scale factor was 1.5 but I'm not sure)
Click to expand...
Q.23 - The answer was £200 if i remember correctly.
Q.24 - This were right at the start of my paper though. I think it ended up being something like 30 for the 2nd half of the question, can't remember what the first half was.
 
georges said:
Thing is.. the easier way for simpler stuff is easier ;)

I only use that formula for the more complex ones on calculator papers. You really don't need it at all for non calc papers.
Click to expand...
I use it for nearly all, unless it's blatantly obvious. Life's too short to spend time working out which pairs of numbers multiply to give one thing and add to give others. Shove the numbers in, get the answers with little or no problems.

Still, I suppose I'm used to getting Complex answers... like solving x^2 + 4x + 5... which works out as 2 +/- i, i think.
 
Beansprout said:
Yes! I don't know why the hell I remember this but it does exist!

I've no idea what exactly it is, though :p
Click to expand...

You mean wave cut platform. Its formed up undercutting of a cliff by hydrolic action. Once enough cliff has undercut it colapses and is washed way, the result is a wave cut platform.
 
csmager said:
I use it for nearly all, unless it's blatantly obvious. Life's too short to spend time working out which pairs of numbers multiply to give one thing and add to give others. Shove the numbers in, get the answers with little or no problems.

Still, I suppose I'm used to getting Complex answers... like solving x^2 + 4x + 5... which works out as 2 +/- i, i think.
Click to expand...

I use it all the time as well, guaranteed result :)
 
csmager said:
I use it for nearly all, unless it's blatantly obvious. Life's too short to spend time working out which pairs of numbers multiply to give one thing and add to give others. Shove the numbers in, get the answers with little or no problems.

Still, I suppose I'm used to getting Complex answers... like solving x^2 + 4x + 5... which works out as 2 +/- i, i think.
Click to expand...

Yeah.. uni is slightly different to gcse higher ;)
 
Angel said:
You mean wave cut platform. Its formed up undercutting of a cliff by hydrolic action. Once enough cliff has undercut it colapses and is washed way, the result is a wave cut platform.
Click to expand...
notch exists too. :P
 
georges said:
Yeah.. uni is slightly different to gcse higher ;)
Click to expand...
Yep, it's awful. Can't wait till next year, no more maths! Some of the stuff really does my head in. Bring back A-Level, all is forgiven!
 
You must log in or register to reply here.
Back
Top Bottom