Maths help

username1490724

How do you find the solution to x^3-27x-90=0.Thanks for the help in advance.

Reply 1

RhymeAsylumForever

Use this formula where a=1, b=0, c=-27 and d=-90

Reply 2

brianeverit

Original post by xpointx

How do you find the solution to x^3-27x-90=0.Thanks for the help in advance.

If this is A-level, then only graphically or numerically
However there is a method which leads to the solution of

x^3-3ax-b=0

being given by

u=\sqrt[3]{\frac{1}{2}(b+\sqrt{b^2+4a^3})}, v=\frac{a}{u}

then solution of equation is u+v

Reply 3

alow

Original post by xpointx

How do you find the solution to x^3-27x-90=0.Thanks for the help in advance.

Is this for FP1?

You can expand the equation:

(x-\alpha)(x-\beta)(x-\gamma)=0

And compare it to your cubic to work out the values of

\alpha, \: \beta \: \text{and} \: \gamma.

That should give you the roots with a bit of manipulation I would think.

(edited 9 years ago)

Reply 4

Sena5

x^3 - 27x - 90 = 0

x^3 - 27x = 90

x ( x^2 - 27) = 90

x ( x^2 - x^3 ) = 90

:. x ( x - 3 ) ( x - 3 ) = 90

So the solutions you can work out.

x=90
x=93

Working

x - 3 = 90
:. x= 90+3
x=93

Reply 5

Notnek

Original post by Sena5

x^3 - 27x - 90 = 0

x^3 - 27x = 90

x ( x^2 - 27) = 90

x ( x^2 - x^3 ) = 90

:. x ( x - 3 ) ( x - 3 ) = 90

So the solutions you can work out.

x=90
x=93

Working

x - 3 = 90
:. x= 90+3
x=93

You've made quite a few mistakes here:

First you changed 27 to x^3 which isn't correct.

Then you said x^2 - x^3 = (x-3)(x-3) which is also wrong.

Then you read off solutions to x ( x - 3 ) ( x - 3 ) = 90 but forgot that the right-hand-side is not 0.