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Mathematics 11 Online
OpenStudy (anonymous):

For what value of x does y=x-x^2 and y=x^2 have the same gradient?

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):

:)

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

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

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

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

OpenStudy (anonymous):

damn

OpenStudy (anonymous):

x = 1 or x=-(1/3)

OpenStudy (anonymous):

yes?

OpenStudy (anonymous):

yes...:)

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