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Maths Question Help

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I've done parts a & b but I'm unsure how to do c, I put dy/dx=0 and then simplified it down to 96x + 128x^3 then solved x to be +0.866 but that's wrong? What have I done wrong can someone help>?
Original post by ForgottenApple
30e91fdb9e151709961b5c1ecd3977f1.png
I've done parts a & b but I'm unsure how to do c, I put dy/dx=0 and then simplified it down to 96x + 128x^3 then solved x to be +0.866 but that's wrong? What have I done wrong can someone help>?


Factorise into x(96+128x^2) and then check for solutions of the quadratic. There is a trivial solution only one value of x.

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(edited 9 years ago)
Original post by ForgottenApple
...


dydx=96x+128x3=0 \displaystyle \frac {dy} {dx} = 96x + 128x^3 = 0 is correct.

Just mistake solving this. Show working?

Good luck. :smile:
Original post by lazy_fish
dydx=96x+128x3=0 \displaystyle \frac {dy} {dx} = 96x + 128x^3 = 0 is correct.

Just mistake solving this. Show working?

Good luck. :smile:


Okay so x=0 or x= root of a negative number so no real roots of that but explain how you know to factorise? I put it in my calculator and it gave me an answer
Original post by ForgottenApple
Okay so x=0 or x= root of a negative number so no real roots of that but explain how you know to factorise?


That's just how you solve equations in general (at least how I do it :tongue:). It's a lot easier to solve a quadratic and a linear seperately than to try and check for all the roots of a cubic.

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davros
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Original post by ForgottenApple
Okay so x=0 or x= root of a negative number so no real roots of that but explain how you know to factorise? I put it in my calculator and it gave me an answer


Because if you ever see an equation like

something = 0

and you can factorize "something" into 2 or more factors you should be doing that automatically since it enables you to see how the individual solutions arise.

(It's the same technique you used at GCSE for simple equations, and it's the same method you use when you have a trig equation and you don't want to lose solutions by dividing by something that might be 0)

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