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**Rose-Red****Member**- Registered: 2006-01-05
- Posts: 19

Could someone help me with this?

x³=6x+9

It doesn't seem difficult, but my answer doesn't suit to the answer in the book

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**God****Member**- Registered: 2005-08-25
- Posts: 59

so basically x^3 - 6x - 9 = 0

We can factor out (x-3) from here, leaving

(x-3)(x^2+3*x+3) = 0

x-3 = 0, so x=3 is a solution

x^2 + 3x + 3 = 0

This can be solved easily using the quadratic formula

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**Rose-Red****Member**- Registered: 2006-01-05
- Posts: 19

thanx man, but I did it in another more complicated way.

x³-9-6x=0

(x³-27)+18-6x=0

(x-3)(x²+27x+27²)-6(x-3)=0

(x-3)(x²+27x+723)=0

x-3=0

x=3

and

x²+27x+723=0

D<0 no solutions

Now the answer is correct.

Could u explain me how you get this (x^2+3*x+3)?

*Last edited by Rose-Red (2006-01-08 09:21:57)*

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**irspow****Member**- Registered: 2005-11-24
- Posts: 455

God factored out x-3 from the original equation. (x-3)(x²+3x+3) = x³-6x-9

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**Rose-Red****Member**- Registered: 2006-01-05
- Posts: 19

but i dont understand how it is possible. I cant find any rule or formula that allows to do that

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**God****Member**- Registered: 2005-08-25
- Posts: 59

By the rational roots theorem, the only possible rational solutions are -9 -3 -1 1 3 and 9

By the rule of signs, there is one sign change, so there is a positive root, so start with positive numbers.

After finding that 3 is a root, we know that we can factor out (x-3)

Use synthetic division to find out the remaining quadratic:

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**Rose-Red****Member**- Registered: 2006-01-05
- Posts: 19

OMG I've never hear of that O_O

Thank you, I'll try to understand that and maybe use it later

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