AS MATHS STRAIGHT LINES / QUADRATICS HELP URGENTLY! Please

Hey so this year I'm doing As maths and I'm stuggling quite a bit. One main reason is my teacher in my old school for GCSE had a very different teaching style and I got A*. My AS maths teacher just prefers us doing it on our own and maybe goes through 1 questions usually the one most people found difficult. (I guess that's what people mean when they say you need to become independent in Alevels) I'm stuggling quite a bit and really need someone to help me by showing me how to do these questions so I can learn from it and do the other ones (I have about 30 questions to do). Any help will be really appreciated and thanks in advance to anyone who does.

Questions

1. The point with coordinates (4p, p^2) lies on the line with the equation
2x - 4y + 5 = 0 .
Find the two possible values of the constant p.

2. Find the coordinates of the points at which each point crosses the coordinates axes.
5x - 3y = 10

3. Solve the simultaneous equation:
3x - y = 49x = 4x^2 + 5

What is the geometric relationship between the two.


4. Use the quadratic formula to solve
4x(x-3) = 11 - 4x

Thanks so much

Q1. (4p, p^2) is the point where x = 4p, y = p^2. So using the equation you have, substitute these values in for x and y and you will get a quadratic in p (I assume these are okay for you given your GCSE grade?)

Q2. This also involves substituting. When a graph crosses the y-axis, the x-coordinate is zero; when it crosses the x-axis, the y-coordinate is zero. Sub in x = 0 and find out what y is, then sub in y = 0 and find out what x is and this will give your two points.

Q3. Not sure what this is meant to be. Is it...
3x - y = 49
x = 4x^2 + 5
Not sure as that would seem like a weird question to me..

Q4.
Expand the brackets, get everything to one side so that you have something equal to zero and apply the formula. If you don't recall it, it can be found by quickly looking it up. (This goes for Q1 too if that doesn't factorise - or if you have trouble factorising)

Hey thanks
sorry about Q.3 it's supposed to be

Solve the simultaneous equation:
3x - y = 4
9x = 4x^2 + 5

Q1. (4p, p^2) is the point where x = 4p, y = p^2. So using the equation you have, substitute these values in for x and y and you will get a quadratic in p (I assume these are okay for you given your GCSE grade?)

Q2. This also involves substituting. When a graph crosses the y-axis, the x-coordinate is zero; when it crosses the x-axis, the y-coordinate is zero. Sub in x = 0 and find out what y is, then sub in y = 0 and find out what x is and this will give your two points.

Q3. Not sure what this is meant to be. Is it...
3x - y = 49
x = 4x^2 + 5
Not sure as that would seem like a weird question to me..

Q4.
Expand the brackets, get everything to one side so that you have something equal to zero and apply the formula. If you don't recall it, it can be found by quickly looking it up. (This goes for Q1 too if that doesn't factorise - or if you have trouble factorising)

Also for q 1 I'm sorry I've forgotten a lot of stuff for GCSE but can u see if it's correct and how do I carry on?

The p term has mysteriously disappeared on the last line. Generally I like to, if there is a -p^2 term, multiply both sides by -1 giving you

4p^2 - 8p - 5 = 0
Now for factorising you've seen that +8 = -2 + 10; likewise you can say here that -8 = 2 - 10

So you can write 4p^2 -10p + 2p - 5 = 0; can you see how to factorise this?
edit: apologies if you saw that silly arithmetic error; I am out of practise

Thanks so much for your time and help! Really do appreciate the help you have been giving me

It's no trouble; I am jobless and awaiting uni, with three months since my exams finished, so I have run short on things to do anyway.

Well thanks how were your results ?

Also can u check out the attachment

Very good thanks, got into my firm. (specifics are on my profile)
Yeah that looks right and the answers also seem to check out with the original equation.

Help

so your only solution for x is x = -1. Sub that into either equation for y to get the corresponding y-value.

I got the answer (-1,4)
It then says for part B to describe the geometrical relationship between the straight line y = 3x + 7 and the curve y = x^2 + 5x +8
??

What do you call a line that just "touches" a curve? As this is what must happen since the line and curve only intersect once (maybe try sketching these to convince yourself).

Yes the curve and straight line meet at the coordinates (-1,4) creating a tangent.
Is that all I should write?

I'd just say something like the line y = 3x + 7 is a tangent to the curve y = x^2 + 5x + 8 (at (-1,4))
What you say is fine but they might be picky and say it doesn't directly describe the relationship

No problem. This is still a weird question as it isn't in the "spirit" of a usual simultaneous equation question but more to the point, simply set the bottom equation equal to zero and then you can solve it like any quadratic (it is factorizable). This will give you two values for x which you can then sub into the top equation for the corresponding y-values.
edit: Phrasing isn't too good there, I mean rearrange the bottom equation so that one side is equal to zero

Sorry if I'm being a pain ! But I'm stuck on this
I literally wanna cry! I've got soo many questions left to do but thanks for your help I probably wouldn't have got through these without you.

It's no problem (though I'll be watching the rugby soon so I cannot help all night ) I just hope that you are getting the general ideas from this questions and I haven't revealed a little too much in some of them.

This one is factorizable; you can find two numbers that multiply to make 20 and add to make -9. However the quadratic formula is always a resource if the factorization isn't jumping out at you.