Expert: Ahmed Salami - 2/3/2010

Question
hi!My question is; x^3+15x+1=0 show that the function has 3 solutions in [-4,4].I know that i should solve it with intermediate value thm but i don not know how i do it.Thanks in advance.

Answer
Hi Buse,
The cubic equation of the form x³ + px + q = 0 has three real and different roots when 4p³ + 27q² < 0 and has one real and two complex roots when 4p³ + 27q² > 0
For x³ + 15x + 1 = 0
4(15)³ + 27(1)² > 0
and so its clear that there is just one real root and two complex roots.
The real root can be found by the Tartaglia's solution.
x = ³√[B + √(B² + A³)] + ³√[B - √(B² + A³)]
where B = -q/2 and A = p/3
For x³ + 15x + 1 = 0
A = 15/3
 = 5
B = -1/2
x = ³√[-½ + √(1/4 + 125)] + ³√[-½ - √(1/4 + 125)]
 = ³√[-½ + √(1/4 + 125)] + ³√[-½ - √(1/4 + 125)]
 = ³√[-½ + 11.1915] + ³√[-½ - 11.19]
 = ³√[10.6915] + ³√[-11.6915]
 = 2.2030 + (-2.2696)
 = -0.0666

Good luck.

Regards