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Maths Qs. Pg 41 Text Test 6
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13-09-2014 01:45PM#1
Comments
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What have you tried so far?
I'll help you with the first and you do the second
x^4 - 9x^2 = 0
We can take out an x squared to give:
x^2(x^2 - 9) = 0
Then inside the brackets it's a difference of two squares:
x^2(x+3)(x-3) = 0
Then that gives us the first point:
x^2 = 0
therefore x = 0
To get the second point
x + 3 = 0
x = -3
And
x - 3 = 0
x = 3
So the answer is x = 0, -3, 30 -
What have you tried so far?
I'll help you with the first and you do the second
x^4 - 9x^2 = 0
We can take out an x squared to give:
x^2(x^2 - 9) = 0
Then inside the brackets it's a difference of two squares:
x^2(x+3)(x-3) = 0
Then that gives us the first point:
x^2 = 0
therefore x = 0
To get the second point
x + 3 = 0
x = -3
And
x - 3 = 0
x = 3
So the answer is x = 0, -3, 3
Thank you so much.0 -
What have you tried so far?
I'll help you with the first and you do the second
x^4 - 9x^2 = 0
We can take out an x squared to give:
x^2(x^2 - 9) = 0
Then inside the brackets it's a difference of two squares:
x^2(x+3)(x-3) = 0
Then that gives us the first point:
x^2 = 0
therefore x = 0
To get the second point
x + 3 = 0
x = -3
And
x - 3 = 0
x = 3
So the answer is x = 0, -3, 3
When I do the second one I get 8x^3-12x^2+6x-1 for the answer of (2x-1)^3
am i on the right track ?0 -
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