x-3-x-16-0-

Question Number 89687 by student work last updated on 18/Apr/20

x^{3} + x - 16 = 0

Commented by mr W last updated on 18/Apr/20

https://en.m.wikipedia.org/wiki/Cubic_equation

Commented by mr W last updated on 18/Apr/20

{i t}^{'} s a l r e a d y s o l v e d i n Q 89653 .

m e t h o d i s a l s o t o l d y o u . h a v e y o u t r i e d

t o s t u d y t h e m e t h o d ? i f y o u j u s t w a n t

t o k n o w t h e f o r m u l a, y o u c a n f i n d i t

i n i n t e r n e t . i f y o u w a n t t o u n d e r s t a n d

i t, y o u s h o u l d s t u d y i t t h r o u g h s o m e

r e a d i n g b y y o u r s e l f .

Commented by MJS last updated on 18/Apr/20

{Cardano}^{'} s Method

{a x}^{3} + {b x}^{2} + c x + d = 0

(1) divide by a \Rightarrow

x^{3} + \frac{b}{a} x^{2} + \frac{c}{a} x + \frac{d}{a} = 0

(2) let z = x - \frac{b}{3 a} \Rightarrow

z^{3} + \frac{3 a c - b^{2}}{3 a^{2}} z + \frac{27 a^{2} d - 9 a b c + 2 b^{3}}{27 a^{3}} = 0

(3) let \frac{3 a c - b^{2}}{3 a^{2}} = p and \frac{27 a^{2} d - 9 a b c + 2 b^{3}}{27 a^{3}} = q \Rightarrow

z^{3} + p z + q = 0 \Leftrightarrow z^{3} = - p z - q

(4) {Cardano}^{'} s new idea : let z = u + v on lhs

{(u + v)}^{3} = - p z - q

u^{3} + 3 u^{2} v + 3 {u v}^{2} + v^{3} = - p z - q

3 u v (u + v) + u^{3} + v^{3} = - p z - q

3 u v z + u^{3} + v^{3} = - p z - q

by matching constants we get

{\begin{cases} I . u v = - \frac{p}{3} \Leftrightarrow u^{3} v^{3} = - \frac{p^{3}}{27} \\ II . u^{3} + v^{3} = - q \end{cases}

(5) let u^{3} = s and v^{3} = t \Rightarrow

{\begin{cases} I . s t = - \frac{p^{3}}{27} \\ II . s + t = - q \end{cases}

this leads to a quadratic :

I . \Rightarrow s = - \frac{p^{3}}{27 t}; insert in II . \Rightarrow

t - \frac{p^{3}}{27 t} = - q \Leftrightarrow t^{2} + q t - \frac{p^{3}}{27} = 0 \Rightarrow

\Rightarrow t = - \frac{q}{2} \pm \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}

\Rightarrow s = - \frac{q}{2} \mp \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}

\Rightarrow we are free to choose the “ +^{″} value

for t and the “ -^{″} value for s

\Rightarrow u = \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} and v = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}

\Rightarrow z_{1} = u + v

z^{3} + p z + q = (z - z_{1}) (z^{2} + (u + v) z + u^{2} - u v + v^{2})

and we can solve this

the only limitation is

D = \frac{q^{2}}{4} + \frac{p^{3}}{27} ⩾ 0

if D < 0 we need the trigonometric solution

Answered by TANMAY PANACEA. last updated on 18/Apr/20

f (x) = x^{3} + x - 16

f (1) = - 14

f (0) = - 16

f (2) = - 6

f (3) = 14

f (4) = 54

n o w f (2) < 0 b u t f (3) > 0

s o o n e r o o t l i e s b e t w e e n 2 a n d 3

3 > x > 2

f (- 1) = - 18

f (- 2) = - 26

s o i t h i n k x^{3} + x - 16 = 0

h a s o n l y o n e s o l u t i o n w h e r e 3 > x > 2