You are Here: Home >< Maths

# Few Higher Qs: algebra, denominator rationalising, simultaneous equation, factorising watch

1. There are a few questions I couldn't figure out, perhaps you can help me out.

1) Here is a number machine:

Input -> multiply by 4 -> subtract 8 -> Output

When the input is a the output is b.

When the input is b the output is c.

Show clearly that c = 8(2a - 5)

I don't understand where the -5 comes from. I thought it would be -40, since c = 16a - 40.

2) I'm rationalising the denominator and simplifying:

77 divided by the square root of 11.

I multiplied it by square root of 11, which creates:

(77 x square root of 11) divided by 11.

Doesn't look simplified to me though, can't figure what I'm supposed to do next.

3) Solve the simultaneous equations:

y = 3x + 1
y squared = 4xsquared (as in 4 times x times x) - x + 7

I know y + ysquared = 2x + 8 + 4xsquared but that's as far as I could go. I don't know how this one's worked out.

4) Factorise 3nsquared (as in 3 x n x n) + 7n + 4

I think I'm in the ballpark with (3n + 7 + 4)(n+1) = 3 x n x n + 3n + 7n + 7 + 4n + 4
2. (1) because you have taken out 8 as a common factor
3. (2) cancel the 77 and the 11
4. (3) square the 3x+1 to give y^2

then put it equal to the expression that you have for y^2
5. (4) not sure which ballpark you are in but that is just a straightforward factorise into 2 brackets question
6. (Original post by ozzyoscy)
I don't understand where the -5 comes from. I thought it would be -40, since c = 16a - 40.
Look at what's outside the brackets.

(Original post by ozzyoscy)
(77 x square root of 11) divided by 11.

Doesn't look simplified to me though, can't figure what I'm supposed to do next.
77 = 7 * 11

(Original post by ozzyoscy)
4)
I think I'm in the ballpark with (3n + 7 + 4)(n+1) = 3 x n x n + 3n + 7n + 7 + 4n + 4
.

3) I can't do off the top of my head.

EDIT: Damn, beaten to it!

Ok for 3 then, like TenOfThem said.
7. (Original post by claret_n_blue)
Look at what's outside the brackets.
Duh, how did I miss that? Yeah I see it.

(Original post by TenOfThem)
(2) cancel the 77 and the 11
(Original post by claret_n_blue)
77 = 7 * 11
7 x (square root of 11 divided by 11). I recognised 77 was a multiple of 11, but I'm not seeing how I can use that.

(Original post by TenOfThem)
(3) square the 3x+1 to give y^2

then put it equal to the expression that you have for y^2
Oh of course. So y^2 = 5x^2 + 6 - x, and y is the square root of that. Does it go further?

(Original post by TenOfThem)
(4) not sure which ballpark you are in but that is just a straightforward factorise into 2 brackets question
(Original post by claret_n_blue)
.

Been doing mostly foundation stuff unfortunately, so it's iffy at the moment. I know how to multiply out brackets, tougher to do it the other way around. I'm getting the hang of where there are two operations, but this one has 3. I guess I have to put a minus somewhere, but knowing that there's just two brackets helps to know as a jumping off point.

Spank you both so far.
8. (Original post by ozzyoscy)

Oh of course. So y^2 = 5x^2 + 6 - x, and y is the square root of that. Does it go further?
Not correct

y^2 = (3x+1)^2

y^2 = 4x^2 - x + 7

therefore

(3x+1)^2 = 4x^2 - x + 7

expand, simplify, solve
9. (Original post by ozzyoscy)
Been doing mostly foundation stuff unfortunately, so it's iffy at the moment. I know how to multiply out brackets, tougher to do it the other way around. I'm getting the hang of where there are two operations, but this one has 3. I guess I have to put a minus somewhere, but knowing that there's just two brackets helps to know as a jumping off point.

Spank you both so far.
The basic "method" when factorising a quadratic is looking for two numbers which multiply to give the last number (in your case its 4) and add to give the middle number (in your case its 7). Also you won't two other numbers that multiply to give you your first number. It's therefore easier if the first number is 1 and not 3 (like you have).

First you needed two numbers that multiply to give you your first number. The only option is 3 * 1 = 3. So right now your brackets are looking like this:

(3x + a)(x + b)

You know need to decided what two numbers multiply to give 4 and add to give 7. Obviously there are none, but don't forget that when expanding these brackets, you will multiply one of these numbers by 3. So now you are looking for two numbers that multiply to give for and two numbers where 3a + b = 7.

Can you think what these numbers are? The easiest way to do it would be to think of the two that multiply to give for and then trial and error from there.
10. How about the (77 x square root of 11) divided by 11 a.k.a. 7 x (square root of 11 divided by 11)? What am I missing there?

(Original post by TenOfThem)
Not correct

y^2 = (3x+1)^2

y^2 = 4x^2 - x + 7

therefore

(3x+1)^2 = 4x^2 - x + 7

expand, simplify, solve
I got (3x+1)^2 which makes 9x^2 + 1. Taking 1 from each side makes:

9x^2 = 4x^2 - x + 6

Take 4x^2 from each side makes:

5x^2 = -x + 6

That's as far as I got, and it doesn't look right.

(Original post by claret_n_blue)
The basic "method" when factorising a quadratic is looking for two numbers which multiply to give the last number (in your case its 4) and add to give the middle number (in your case its 7). Also you won't two other numbers that multiply to give you your first number. It's therefore easier if the first number is 1 and not 3 (like you have).

First you needed two numbers that multiply to give you your first number. The only option is 3 * 1 = 3. So right now your brackets are looking like this:

(3x + a)(x + b)

You know need to decided what two numbers multiply to give 4 and add to give 7. Obviously there are none, but don't forget that when expanding these brackets, you will multiply one of these numbers by 3. So now you are looking for two numbers that multiply to give for and two numbers where 3a + b = 7.

Can you think what these numbers are? The easiest way to do it would be to think of the two that multiply to give for and then trial and error from there.
Oh yeah I remember when I first tried, I knew you had to get two numbers that added together to make one number and multiplied to make another, but like you said that doesn't work with 4 and 7 so I was stumped.

From what you've said I've found it's (3n + 4)(n + 1)

I was in the ballpark! I just had to remove the 7 I put in the first bracket.
11. (Original post by ozzyoscy)
How about the (77 x square root of 11) divided by 11 a.k.a. 7 x (square root of 11 divided by 11)? What am I missing there?

I got (3x+1)^2 which makes 9x^2 + 1.
No it does not = that
12. (Original post by TenOfThem)
I'll have to remember all those division/multiplication substitution tricks. Thanks.

No it does not = that
3 x 3 = 9
x x x = x^2
1 x 1 = 1
Thus (3x + 1)^2 = 9(x^2) +1, right?
13. (Original post by ozzyoscy)

3 x 3 = 9
x x x = x^2
1 x 1 = 1
Thus (3x + 1)^2 = 9(x^2) +1, right?
no

(3x+1)^2 = (3x+1)(3x+1)
14. (Original post by TenOfThem)
no

(3x+1)^2 = (3x+1)(3x+1)
Gah, maths is funny sometimes.

3x x 3x = 9x^2
3x x 1 = 3x
1 x 3x = 3x
1 x 1 = 1

(3x+1)^2 = 9x^2 + 6x + 1

So now I have:

y^2 = 9x^2 + 6x + 1
y^2 = 4x^2 - x + 7

Then I remove like terms from both sides, and I have y^2 = 5x^2 + 6 + 5x
15. (Original post by ozzyoscy)
Gah, maths is funny sometimes.

3x x 3x = 9x^2
3x x 1 = 3x
1 x 3x = 3x
1 x 1 = 1

(3x+1)^2 = 9x^2 + 6x + 1

So now I have:

y^2 = 9x^2 + 6x + 1
y^2 = 4x^2 - x + 7
This is correct

Then I remove like terms from both sides, and I have y^2 = 5x^2 + 6 + 5x

This makes no sense, I cannot see what you are doing

y^2=y^2
16. (Original post by TenOfThem)
This makes no sense, I cannot see what you are doing

y^2=y^2
y^2 = 9x^2 + 6x + 1 = 4x^2 - x + 7

9x^2 - 4x^2 = 5x^2, 6x - x = 5x, 7 - 1 = 6

So y^2 = 5x^2 + 5x + 6
17. (Original post by ozzyoscy)
y^2 = 9x^2 + 6x + 1 = 4x^2 - x + 7
This is correct so you have

Which you need to solve to find x

9x^2 - 4x^2 = 5x^2, 6x - x = 5x, 7 - 1 = 6

So y^2 = 5x^2 + 5x + 6
This is nonsense
18. (Original post by TenOfThem)
This is correct so you have

Which you need to solve to find x

This is nonsense
Yikes, no need to be a **** about it. If removing like terms isn't one of the processes to finding the answer, then I don't know the answer.
19. (Original post by ozzyoscy)
Yikes, no need to be a **** about it. If removing like terms isn't one of the processes to finding the answer, then I don't know the answer.
OK

I am sure someone else will be along to help
20. (Original post by TenOfThem)
OK

I am sure someone else will be along to help
http://www.fisme.science.uu.nl/toepa...wisweb.en.html

### Related university courses

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: January 5, 2013
Today on TSR

### Exam Jam 2018

Join thousands of students this half term

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