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How do I rearrange this ?

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Reply 20
Avatar for Tizmo
Tizmo
Original post by eliotball
No problem, I'm sure you had good intentions posting that solution, it was just a pretty awful mistake :P


Um yeah, big mistake! I feel stupid now, haha.
Reply 21
Avatar for birduk
birduk
Original post by kingkongjaffa
thanks how do you come to that? I'm not very good at factorising with things that have coefficients of x :P


You have to think that you have

3x^2 -10x +7 = 0

So something x times x gives 3x^2, so that something can only be 3. That gives you a starting point with two numbers that are unknown (we'll call them a and b)
(3x + a)(x + b)= 0
You also know that if you seperate out this bracket (which I am sure you can do)
You get...
3x^2 + ax + 3bx + ab = 0

So a times be has to be 7. Leaves you with a choice of 1, -1, 7 and -7 for a and b.
You also know that ax + 3bx is -10x, so then you know that a + b is -10. There is only one solution of the a and bs that we have shown will work.

Hope that helps!
Original post by birduk
You have to think that you have

3x^2 -10x +7 = 0

So something x times x gives 3x^2, so that something can only be 3. That gives you a starting point with two numbers that are unknown (we'll call them a and b)
(3x + a)(x + b)= 0
You also know that if you seperate out this bracket (which I am sure you can do)
You get...
3x^2 + ax + 3bx + ab = 0

So a times be has to be 7. Leaves you with a choice of 1, -1, 7 and -7 for a and b.
You also know that ax + 3bx is -10x, so then you know that a + b is -10. There is only one solution of the a and bs that we have shown will work.

Hope that helps!


thanks a lot :smile:
Original post by kingkongjaffa
ooh can you explain this please firstly AC is 3*7 but can you explain why you did that.
secondly can you explain how you go from 3x(x-1) -7(x-1) to (3x-7) (x-1)?


Well because the co-efficient of x^2 is greater than two, you must multiply the co-efficient of x^2 by the constant which is A x C ( if confused as to the letters, remember the quadratic equation).

You go from 3x(x-1) -7(x-1) to (3x-7) (x-1) by eliminating one (x-1) and making the numbers outside each bracket into another bracket. REMEMBER to eliminate one bracket, both brackets have to be the same.....

eg: 4x(x+5) 5(x+5) will become (4x +5)(x+5)

and, 11x(x-3) -4(x-3) = (11x-4)(x-3)


Note the Solutions for x exist when you make the bracket equal to 0...

eg (4x + 5)(x+5) =0

4x+5=0
4x= -5
x= -5/4

and, x +5 = 0
x = -5
Reply 24
Avatar for eliotball
eliotball
7
Original post by Tizmo
Um yeah, big mistake! I feel stupid now, haha.


The feeling will wear off :cool:
Original post by crazycake93
Well because the co-efficient of x^2 is greater than two, you must multiply the co-efficient of x^2 by the constant which is A x C ( if confused as to the letters, remember the quadratic equation).

You go from 3x(x-1) -7(x-1) to (3x-7) (x-1) by eliminating one (x-1) and making the numbers outside each bracket into another bracket. REMEMBER to eliminate one bracket, both brackets have to be the same.....

eg: 4x(x+5) 5(x+5) will become (4x +5)(x+5)

and, 11x(x-3) -4(x-3) = (11x-4)(x-3)


Note the Solutions for x exist when you make the bracket equal to 0...

eg (4x + 5)(x+5) =0

4x+5=0
4x= -5
x= -5/4

and, x +5 = 0
x = -5



Thanks i understand it now the examples helped aswell :smile: thanks for taking the time to read the thread, all of you.:smile:

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