You are Here: Home >< Maths

# Gcse help Watch

1. i dont want you to just do the question but could you actually explain how you're doing it. that'd be great.
find the value of x for which:
2x^2 - 7x + 3 > 0

Thanks if you could help..... i just dont know what to do when there is a squared in there! :|
2. Factorise?
3. (Original post by boromir9111)
Factorise?
ok i know what factorising it but i dont knw how you'd factorise this?! im really bad at maths !
4. two numbers that multiply to gove 3 (the constant term at the end) and add to give 14 (it is 14 because 2 x 7 [7 is from the middle]. and the 2 comes from the 2 in front of the x^2)

I don't know how you're going to remember it, so just make up some weird mnemonic for it.

the answer is (2x - 1) (x-3)

so x= 1/2, 3.

do you want me to explain anything else you're unclear about? quote me so I see reply
5. (Original post by ilyking)
two numbers that multiply to gove 3 (the constant term at the end) and add to give 14 (it is 14 because 2 x 7 [7 is from the middle]. and the 2 comes from the 2 in front of the x^2)

I don't know how you're going to remember it, so just make up some weird mnemonic for it.

the answer is (2x - 1) (x-3)

so x= 1/2, 3.

do you want me to explain anything else you're unclear about? quote me so I see reply
Thanks for that! i understand it now
i am now stuck on another question! GRAPHS....... i hate them
it says:
the points A and B have co-ordinates (1, 2) and (5, 8)
find, in the form y=mx+c an equation for the line through both points A and B
...i dont have a clue!
6. (Original post by Jay1012)
Thanks for that! i understand it now
i am now stuck on another question! GRAPHS....... i hate them
it says:
the points A and B have co-ordinates (1, 2) and (5, 8)
find, in the form y=mx+c an equation for the line through both points A and B
...i dont have a clue!
What is m? how do you find it? what is c? how do you find it?
7. (Original post by boromir9111)
What is m? how do you find it? what is c? how do you find it?
m and c are just numbers i think....
you have to find an equation of the line that goes through those 2 points and just give it in the form y=mx+c m and c are just integers
8. (Original post by Jay1012)
Thanks for that! i understand it now
i am now stuck on another question! GRAPHS....... i hate them
it says:
the points A and B have co-ordinates (1, 2) and (5, 8)
find, in the form y=mx+c an equation for the line through both points A and B
...i dont have a clue!
hmmm you seem to have a good understanding so I'll try and explain it through tsr

you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

m (gradient) = change in y / change in x so.....

m = 2 - 8 / 1 - 5 = -3/2

then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

formula is: y - y1 = m(x-x1) <<<(equation of line formula, in order to use this we need to have ONE coordiante and a gradient [which we just figured out to be -3/2]

which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

y - 2 = -3/2 (x-1)

so... y = -3/2x + 7/2

and done.

seem long, but not really...
9. (Original post by ilyking)
hmmm you seem to have a good understanding so I'll try and explain it through tsr

you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

m (gradient) = change in y / change in x so.....

m = 2 - 8 / 1 - 5 = -3/2

then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

formula is: y - y1 = m(x-x1)

which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

y - 2 = -3/2 (x-1)

so... y = -3/2x + 7/2

and done.

seem long, but not really...
You know the forum rules. Don't post full solutions!
10. (Original post by boromir9111)
You know the forum rules. Don't post full solutions!
oh yeah and my name is aragorn
11. (Original post by ilyking)
oh yeah and my name is aragorn
I had to +rep you for that even though you broke the forum rules
12. (Original post by ilyking)
hmmm you seem to have a good understanding so I'll try and explain it through tsr

you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

m (gradient) = change in y / change in x so.....

m = 2 - 8 / 1 - 5 = -3/2

then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

formula is: y - y1 = m(x-x1) <<<(equation of line formula, in order to use this we need to have ONE coordiante and a gradient [which we just figured out to be -3/2]

which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

y - 2 = -3/2 (x-1)

so... y = -3/2x + 7/2

and done.

seem long, but not really...
thanks! youre great help! i understand what you're telling me but it's just knowing what to do to solve to question!
i hate Graph stuff, anything to do with graphs im awful at.
Could you explain this one last question.... maybe just the steps to take and start me off

The line y=41 meets C at point R
find the x coordinate of R, giving your answer in the form p+q(square root of 2) where p and q are integers..... so surd form i think?
13. (Original post by ilyking)
hmmm you seem to have a good understanding so I'll try and explain it through tsr

you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

m (gradient) = change in y / change in x so.....

m = 2 - 8 / 1 - 5 = -3/2

then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

formula is: y - y1 = m(x-x1) <<<(equation of line formula, in order to use this we need to have ONE coordiante and a gradient [which we just figured out to be -3/2]

which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

y - 2 = -3/2 (x-1)

so... y = -3/2x + 7/2

and done.

seem long, but not really...
another question im stuck on........
the curve C with equation y=x^2 - 6x + 18 meets the y-axis at P and has a minimum point at Q.
sketch the graph of C, showing the coordinates of P and Q...
how would you work these 2 points out and what would the graph look like?
thanks if you could help....... i wont bother you after this one

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 30, 2010
Today on TSR

### Cambridge interview invitations

Has yours come through yet?

### Official Oxford interview invite list

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.