Maths question

mufc802

mufc802

mufc802

Soldato
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1 Jul 2009
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Got the equations y=(x-1)^3 and y=(x-1)(x+1)

Drawn them and they intersect each other 3 times. So I need to find the y co-ordinates of where they intersect.

I've got:
(x - 1)^3 = (x-1)(x+1)
(x - 1)(x - 1)(x - 1) = x^2 - 1
(x - 1)(x^2 - 2x + 1) = x^2 - 1
x^3 - 2x^2 + x - x^2 + 2x - 1 = x^2 - 1
x^3 - 4x^2 + 3x = 0
x(x^2 - 4x + 3) = 0
Can't seem to factorise this..
Get x(x^2 - 3x - x + 3) = 0
x[x(x-3)-1(x+3)]
x(x-1)(x-or+3?)

Might have missed something obvious.:o
 
x = 3, 0, 1

Via the above.

You can get two of the solutions very quickly by eliminating (x - 1) from both sides at the beginning, which would've got you the + or - 3 you were looking for.
 
Hxc said:
x = 3, 0, 1

Via the above.

You can get two of the solutions very quickly by eliminating (x - 1) from both sides at the beginning, which would've got you the + or - 3 you were looking for.
Click to expand...

Do you mean (x - 1)^3 = (x-1)(x+1)? Can I just eliminate (x-1) so would it become (x-1)^2 = (x+1)?
 
You can do that but you'll lose a solution, x = 1. It will get you the other two very quickly and x=1 can be seen very easily in this case from the equasions at hand so that's the way I would do it.

Alternatively in an A level exam (presuming that's what this is?) I would put them in my graphic calculator and make it do it for me :)
 
So you have, essentially:

(x-1)^3 - (x+1)(x-1) = 0
It is generally not good practice to divide through by (x-1), since you are potentially eliminating a solution, and you could be dividing through by zero. Take it out as a factor instead.
So now:
(x-1)[(x-1)^2 - (x+1)] = 0
(x-1)[x^2 - 2x + 1 - x - 1] = 0
(x-1)[x^2 - 3x] = 0
(x-1)[x(x-3)] = 0, or a more clean way of writing it:
x(x-1)(x-3) = 0, giving you your three roots immediately by observation :)
 
Mr Jack said:
x^2 -4x + 3 = (x-3)(x-1) if that helps.
Click to expand...

BTW, in general if you have an equation of the form x^2 + Ax + B you are looking for a solution (x + c)(x + d) where c+d = A and cd = B, so in this case you need two numbers that multiply to 3, which can only be 3, 1 or -3,-1 and you can easily see that it solves to -3, -1 because 3+1 = 4 and you need -4.

(Obviously this only works if they're kind enough to use integer co-efficients but they generally are)
 
Mr Jack said:
BTW, in general if you have an equation of the form x^2 + Ax + B you are looking for a solution (x + c)(x + d) where c+d = A and cd = B, so in this case you need two numbers that multiply to 3, which can only be 3, 1 or -3,-1 and you can easily see that it solves to -3, -1 because 3+1 = 4 and you need -4.

(Obviously this only works if they're kind enough to use integer co-efficients but they generally are)
Click to expand...

I'm usually fine with this sort of thing, just got my - and + mixed up..

I just get the equation ax^2 + bx + c and firstly do ac if necessary, then do ax^2 + bx + bx + ac and the two bx add to the first bx and multiply to make ac. No guesswork required but might take longer.
 
Here's my attempt :)

210wu1h.jpg


Forgot to add a plus sign in the binomial expansion at the top, but apart from that I think that's it
 
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