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1. I've been given this question by a friend and I'm stuck on it.

The roots of the equationare a and a+4. Find the possible values of p.

I started by expanding the bracket but I have no idea if that's right. I can sort of see what it's getting at though.

What confused me is the use of the letter a, since the formula for quadratics is ax^2..., I thought a=1 but I don't think it would be that easy...

Edit: This is what I have now:

Not sure what to do from here

Would I use x=10?
2. x^2 + (p-10)x - 10p = 0 is how it would look in your more 'general' form I presume you are use to.
here the roots are x = a and x = a+4

Put these in to the equation one at a time and solve for p (remember you are trying to find p not a)
3. (Original post by Bude8)
I've been given this question by a friend and I'm stuck on it.

The roots of the equationare a and a+4. Find the possible values of p.

I started by expanding the bracket but I have no idea if that's right. I can sort of see what it's getting at though.

What confused me is the use of the letter a, since the formula for quadratics is ax^2..., I thought a=1 but I don't think it would be that easy...

Edit: This is what I have now:

Not sure what to do from here

Would I use x=10?
Taking out the common factor gives you (x+p)(x-10)
4. (Original post by TenOfThem)
Taking out the common factor gives you (x+p)(x-10)
Wait, how did you get that?
5. (Original post by Bude8)
I've been given this question by a friend and I'm stuck on it.

The roots of the equationare a and a+4. Find the possible values of p.

I started by expanding the bracket but I have no idea if that's right. I can sort of see what it's getting at though.

What confused me is the use of the letter a, since the formula for quadratics is ax^2..., I thought a=1 but I don't think it would be that easy...

Edit: This is what I have now:

Not sure what to do from here

Would I use x=10?
Use the Viéte formula
For ax^2+bx+c=0 where the two roots are , then

From above for example

a=1 b=p-10 c=-10p and (x_1-x_2)^2=16
6. (Original post by Bude8)
Wait, how did you get that?

x(x-10) + p(x-10)

Common factor
7. (Original post by TenOfThem)

x(x-10) + p(x-10)

Common factor
So the (x-10) cancel out, but aren't you left with just x+p?
I don't get why there's still a (x-10)
I probably need to revise factorising again...
8. (Original post by Bude8)
So the (x-10) cancel out, but aren't you left with just x+p?
I don't get why there's still a (x-10)
I probably need to revise factorising again...
I do not understand how you think they "cancel out"

It is like

ax+bx = x(a+b)

The x doesn't "cancel"
9. (Original post by TenOfThem)
I do not understand how you think they "cancel out"

It is like

ax+bx = x(a+b)

The x doesn't "cancel"
ok so from x(x-10)+p(x-10)=0
you get
(x+p)(x-10)

What I don't understand is why one of the (x-10) goes...
So confused right now
10. (Original post by Bude8)
ok so from x(x-10)+p(x-10)=0
you get
(x+p)(x-10)

What I don't understand is why one of the (x-10) goes...
So confused right now
since if you take out a common factor of (x-10) you'll notice that you times this by x and p which the same as (x+p)(x-10), basically if you look there you get p(x-10) and x(x-10), look at the bracket and see why it makes sense

It's not gone, you've just factored it by using foil or otherwise you can see that (x+p)(x-10) gives the exact same result
11. (Original post by Robbie242)
since if you take out a common factor of (x-10) you'll notice that you times this by x and p which the same as (x+p)(x-10), basically if you look there you get p(x-10) and x(x-10), look at the bracket and see why it makes sense
Ah, I see, thank you!

Rep for everyone who has helped, I've solved it now.
12. (Original post by Bude8)
ok so from x(x-10)+p(x-10)=0
you get
(x+p)(x-10)

What I don't understand is why one of the (x-10) goes...
So confused right now
Then I agree

You need to revise factorising/multiplication of brackets

In your first post you went from (p-10)x to px - 10x
Did you wonder where the "extra" x came from
This is the same question!
13. (Original post by Bude8)
Wait, how did you get that?
the distributivity

(x-10)[A+B]=A(x-10)+B(x-10)

in reverse order is a factorizing
A(x-10)+B(x-10)=(x-10)(A+B)
14. (Original post by Bude8)
Ah, I see, thank you!

Rep for everyone who has helped, I've solved it now.
good

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