Mathematics
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OpenStudy (anonymous):
For what value of x does y=x-x^2 and y=x^2 have the same gradient?
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OpenStudy (anonymous):
differentiate each and equate the derivatives
OpenStudy (anonymous):
x=1/4
OpenStudy (anonymous):
woah @akash123 where'd that come from?
OpenStudy (anonymous):
differentiate each and equate the derivatives
OpenStudy (anonymous):
:)
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OpenStudy (anonymous):
You were scaring me there
OpenStudy (anonymous):
@JJ1371 taking it in...
OpenStudy (anonymous):
where? proving the collinearity..
OpenStudy (anonymous):
no, just suddenly pulling an answer
OpenStudy (anonymous):
oh...ok
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OpenStudy (anonymous):
sorry, the second y=x^3
OpenStudy (anonymous):
Ok, follow the procedure
OpenStudy (anonymous):
y=1+2x
y=2x^2
1+2x = 2x^2
where to now?
OpenStudy (anonymous):
now u have to solve this quadratic equation and find the values of x
OpenStudy (anonymous):
1+2x = 3x^2
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OpenStudy (anonymous):
okay
2x^2-2x-1=0
OpenStudy (anonymous):
derivative of x^3 = 3 x^2
OpenStudy (anonymous):
how did you get that?
OpenStudy (anonymous):
derivative of x^n = n x^(n-1)
OpenStudy (anonymous):
oh right yeah
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OpenStudy (anonymous):
sorry
OpenStudy (anonymous):
3x^2 - 2x -1 = 0
Quadratic equation?
OpenStudy (anonymous):
wait!
(x-3)(x+1) = 0
OpenStudy (anonymous):
therefore x = 3 or x = -1?
OpenStudy (anonymous):
no...3x^2 - 2x -1 = 0 =(x-3)(x+1) is wrong
do the factorization again
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OpenStudy (anonymous):
damn
OpenStudy (anonymous):
x = 1
or
x=-(1/3)
OpenStudy (anonymous):
yes?
OpenStudy (anonymous):
yes...:)