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Maths Correction. - HSN forum

# Maths Correction.

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### #1Rocky

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Posted 08 September 2005 - 07:03 PM

*Can you help of possible for this Q.

1- If the points (3, -1), (a, 2) and (b, 5) are collinear show that
"2a -b = 3"

Am I right for this Q?
2- f(x)= x squared - 3 and g(x)= 2x + 1
(a) Find f( g(x) ) and g( f(x) )

* I got f( g(x) )= 2x squared - 5
And g( f(x) )= 2X squared + 4x - 2
So am i right?

(b) If f( g(x) ) - g( f(x) ) = 9
Find the possible values of x

*It says possibles values but I got "x= 4".
So am i right, or did I do a msitake?

### #2Dave

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Posted 08 September 2005 - 07:44 PM

f(g(x)) = (2x+1)2 - 3
g(f(x)) = 2x2 - 5

see why i got these answers and if you still dont get it come back

If i am not here i am somewhere else

### #3Rocky

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Posted 08 September 2005 - 08:39 PM

How do you then get the x-values though?

### #4Dave

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Posted 08 September 2005 - 09:09 PM

solve the equation normally

(2x+1)2 - 3 - [ 2x2 - 5] = 9

If i am not here i am somewhere else

### #5dfx

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Posted 08 September 2005 - 09:15 PM

The question says:

If f(g(x)) - g(f(x)) = 9

So, now that you have those combined functions, just plug them in:

(2x+1) - 3 - ( 2x - 5) = 9

(2x+1) - 3 - 2x + 5 = 9

4x + 4x + 1 - 3 - 2x + 5 = 9

2x + 4x + 1 - 3 + 5 - 9 = 0

2x + 4x - 6 = 0

Now just proceed to solve the quadratic equation with your favourite method - in this case, splitting the middle term and inspecting:

2x + 6x - 2x - 6 = 0

2x (x + 3) - 2(x + 3) = 0

(2x - 2) (x + 3) = 0 x = 1 OR x = -3

### #6Rocky

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Posted 09 September 2005 - 12:23 PM

Thanx dfx for helping me understand what you did. Also I know how you do collinear Qs but not one where you have to solve to get an equation. So can you help please,
Q- If the points (3, -1), (a, 2) and (b, 5) are collinear show that
"2a -b = 3"

### #7dfx

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Posted 09 September 2005 - 12:36 PM

Sure, but do give it a go yourself. It is really quite simple - just algebra.

Remember, when you're showing collinearity you equate the gradients!

Other than that, there is a very simple trick at the end that might put you off track - you have to simplify *slightly* abstractly. Post back if you don't get it...

Oh and, I did get teh equations from Dave in the first place so you know.

### #8Rocky

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Posted 09 September 2005 - 12:41 PM

I know but you worked out so I appreciated that. I tried the collinearity but it's not working out dude. So if you have time, could you help me please

### #9dondon

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Posted 09 September 2005 - 12:48 PM

you work out the gradient for the first coordinate to the 2nd then the 2nd to the 3rd. And since it is collinear these are equal and on the top of both of the equations is a 3 so the bottom of the equations are equal so
a-3=b-a
2a-b=3

### #10Rocky

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Posted 09 September 2005 - 01:00 PM

How about explain what you did before the last 2 lines lol as I am confused

### #11dondon

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Posted 09 September 2005 - 01:07 PM

ok

Q- If the points (3, -1), (a, 2) and (b, 5) are collinear show that
"2a -b = 3"

First work out the gradient of the first two coordinates

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

Then work out the gradient between the 2nd and 3rd point.

y -y / x -x
=5-2/b-a
=3/b-a

so since the lines are collinear both equations equal one another, but there is a three on top of both of the equations so that is cancelled out so:
a-3=b-a

2a-b=3

### #12Rocky

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Posted 09 September 2005 - 01:23 PM

Thanx dondon

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