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TalkOrigin
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Hi, I'm stuck on what should be an easy problem and I'm not sure where I'm going wrong. I already know the answer (I'm self studying), it's the method for the solution I don't understand. Here's the problem:
a) on the same axes sketch the curves given by $$y= (x-1)^3$$ and
$$y= (x-1)(1+x)$$
b) find the coordinates of the points of intersection
So, I did part a, and I thought i did part b, but I only got 1 of 3 possible solutions.
Anyway, here was my attempt at a solution:
$$ (x-1)^3 = (x-1)(1+x)$$
$$(x-1)^2 = (1+x)$$
Then multiply out the brackets: $$x^2 - 2x + 1 = 1+x$$
Simplify: $$x^2 - 3x = 0$$
Factorise: $$x(x-3) = 0$$
From this I can get 2 values for x (0 and 3) and plug them into the original equations. I know these two are correct, but I'm missing one and I'm pretty sure my method is wrong.
I had a look at the solution (the textbook I have comes with online worked solutions) but I've been staring at it and don't understand it. the solution they give is:
$$(x-1)^3 = (x-1)(x+1)$$
$$(x-1)^3 - (x-1)(x+1) = 0$$
This is the part I don't understand at all:
$$(x-1)[x^2-2x+1-(x+1)] = 0$$
Then $$(x-1)(x^2-3x)=0$$
$$(x-1)(x-3)x = 0$$
$$x = 0, 1, 3$$
Any help on this would be greatly appreciated, thank you.
a) on the same axes sketch the curves given by $$y= (x-1)^3$$ and
$$y= (x-1)(1+x)$$
b) find the coordinates of the points of intersection
So, I did part a, and I thought i did part b, but I only got 1 of 3 possible solutions.
Anyway, here was my attempt at a solution:
$$ (x-1)^3 = (x-1)(1+x)$$
$$(x-1)^2 = (1+x)$$
Then multiply out the brackets: $$x^2 - 2x + 1 = 1+x$$
Simplify: $$x^2 - 3x = 0$$
Factorise: $$x(x-3) = 0$$
From this I can get 2 values for x (0 and 3) and plug them into the original equations. I know these two are correct, but I'm missing one and I'm pretty sure my method is wrong.
I had a look at the solution (the textbook I have comes with online worked solutions) but I've been staring at it and don't understand it. the solution they give is:
$$(x-1)^3 = (x-1)(x+1)$$
$$(x-1)^3 - (x-1)(x+1) = 0$$
This is the part I don't understand at all:
$$(x-1)[x^2-2x+1-(x+1)] = 0$$
Then $$(x-1)(x^2-3x)=0$$
$$(x-1)(x-3)x = 0$$
$$x = 0, 1, 3$$
Any help on this would be greatly appreciated, thank you.