Urgent help needed on Maths question - A level
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Original post by AQASUX
But then when i come to do the simultaneous equation i get the two as:
-75 = 4p + q
and 3x^2 +p
-75 = 4p + q
and 3x^2 +p
Original post by AQASUX
People were telling me to do simultaneous equations before.
So,
y= x^3 + 48x + q
-11 = 64 + 192 + q
245 = q
y=x^3 + 48x + 245
dy/dx = 3x^2 + 48
0 = 3x^2 + 48
0=x^2 + 16
using quadratic formula x = +/- squareroot4
So,
y= x^3 + 48x + q
-11 = 64 + 192 + q
245 = q
y=x^3 + 48x + 245
dy/dx = 3x^2 + 48
0 = 3x^2 + 48
0=x^2 + 16
using quadratic formula x = +/- squareroot4
You have lost me
you had 2 simultaneous equations
-75 = 4p + q
0 = 48 + p
These give p = -48
And -75 = 4*-48 + q so q = .....
Then you have your 2 values of x, one gave the minimum in the OP so use the other for the maximum
(edited 9 years ago)
Original post by TenOfThem
You have lost me
you had 2 simultaneous equations
-75 = 4p + q
0 = 48 + p
These give p = -48
And -75 = 4*-48 + q so q = .....
Then you have your 2 values of x, one gave the minimum in the OP so use the other for the maximum
you had 2 simultaneous equations
-75 = 4p + q
0 = 48 + p
These give p = -48
And -75 = 4*-48 + q so q = .....
Then you have your 2 values of x, one gave the minimum in the OP so use the other for the maximum
q = -267
so then put that into the original
y=x^3-48x-267
the derivative= x^2-48
at turning point
0=x^2-48
48=x^2
x=squareroot48
Original post by AQASUX
q = -267
so then put that into the original
y=x^3-48x-267
the derivative= x^2-48
at turning point
0=x^2-48
48=x^2
x=squareroot48
so then put that into the original
y=x^3-48x-267
the derivative= x^2-48
at turning point
0=x^2-48
48=x^2
x=squareroot48
how do you have 2 values for x ?
Original post by AQASUX
q = -267
No
the derivative= x^2-48
No
With regards to the 2 answers, you had the 2 answers in post 20
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