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[hushed stirrup]

Explain why x³-27=0 has only one real solution

[fair quail]

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[timid ice]

Which real solution is it ?

[hushed stirrup]

No idea

[timid ice]

maybe factor 27

(there are multiple ways to show it has only one real solution, one algebraic and another using only analysis, since you've tagged it as algebra I propose to help you factor the polynomial)

[hushed stirrup]

I think it's quadratic formula

Since the square root will be negative on the inside

[timid ice]

which quadratic formula ? applied to which polynomial ?

[hushed stirrup]

X²+3x-27

[timid ice]

where does it come from ?

[hushed stirrup]

X³-27=(x-3)(x²+3x+9)

[timid ice]

yes, this is it

the roots are either 3 (root of x-3) or a root of (x²+3x+9), and the discriminant Δ is negative, so no real root for this one

[hushed stirrup]

Ty

[gleaming steeple]

because the other solutions are not real

[plain patio]

The function x—>x^3 is injective on R

So there is one real solution and it’s unique

[frosty zinc]

factor into (x-3)(x^2+3x+9)=0 and use quadratic formula on x^2+3x+9 to prove that has no solutions

thus only solution is when x-3=0 meaning x=3

[celest maple]

to be (very) general and blunt, an equation in the form of x^3-a=0 has only 1 real solution, namely that being cbrt(a) (this is because of the factoring and roots shown above)

[elfin prairie]

Wait, so are you talking about x^3 - 27 = 0 or x^3 + 3x - 27 = 0?

[celest maple]

to build onto this, in general, any formula with equation x^n-a=0, where n>2, will always have some sort of imaginary root

i could prove this using the roots of unity, if you want

[elfin prairie]

Why not n = 2, too?

[celest maple]

becuase it doesn’t work?

it’s not like x^2-a can have more than 2 roots

so the only roots are sqrt(a) and -sqrt(a)

[timid ice]

(with the condition that a > 0)

[celest maple]

$a^{\frac{1}{n}}\left(\cos\left(\frac{2\pi k}{n}\right)+i \sin\left(\frac{2\pi k}{n}\right)\right)$

[mortal fern]

;( | 追放された興奮

[celest maple]

where $k\in \mathrm{Z}$ and $n\in \mathrm{N}$

[mortal fern]

;( | 追放された興奮

[celest maple]

thus for n= 1,2 (assuming that a>0) the imaginary part is sin(2pik) and sin(pik)

for these two cases, for all integers, sin would end up being zero, so thus for n=1,2, there are no imaginary solutions if a>0, as presented in this case