Is there another way to solve this polynomial of degree 3?

[ionlydothemath]

I need to solve for x -

log₁₅ (x) + log₁₅ (x - 2) + log₁₅ (x + 3) = 1

this is my attempt so far -

log₁₅ (x)(x - 2)(x + 3) = 1

(x)(x - 2)(x + 3) = 15

x³ + x² - 6x = 15

x³ + x² - 6x - 15 = 0

I can then only manage to solve the polynomial by using this equation -

x = {q + [q² + (r-p²)³]^1/2}^1/3 + {q - [q² + (r-p²)³]^1/2}^1/3 + p

where:

p = -b/(3a)

q = p³ + (bc-3ad)/(6a²)

r = c/(3a)


This gives me 2.88395... which is the correct answer (i believe).

I'm just wondering if there is another way of solving the problem that I might be missing?

Any help would be much appreciated.

 

[Deleted member 4993]

I need to solve for x -

log₁₅ (x) + log₁₅ (x - 2) + log₁₅ (x + 3) = 1

this is my attempt so far -

log₁₅ (x)(x - 2)(x + 3) = 1

(x)(x - 2)(x + 3) = 15

x³ + x² - 6x = 15

x³ + x² - 6x - 15 = 0

I can then only manage to solve the polynomial by using this equation -

x = {q + [q² + (r-p²)³]^1/2}^1/3 + {q - [q² + (r-p²)³]^1/2}^1/3 + p

where:

p = -b/(3a)

q = p³ + (bc-3ad)/(6a²)

r = c/(3a)


This gives me 2.88395... which is the correct answer (i believe).

I'm just wondering if there is another way of solving the problem that I might be missing?

Any help would be much appreciated.

The given function does not have any rational root, So only other way to approximate the roots of this equation is by numerical method - e.g. Newton-Raphson method.