Question

-x^3+3x+1=0
@phi can you help me with this too please

Answer 1

There is a formula for the exact answer, and it looks very weird. I don't remember it.

Answer 2

are you sure you have a cubic ? they make quadratics look trivial

Answer 3

yes

Answer 4

i used a calculator on google and the answer looked really weird had inside a sqrt fractions multiplying

Answer 5

this is a "depressed cubic" because it does not have an x^2 term. We can try
https://en.wikipedia.org/wiki/Cubic_equation#Vieta.27s_substitution

Answer 6

where did the w come from in that equation?

Answer 7

you should multiply through by -1 to get
x^3 -3x - 1 = 0
and match to
x^3 + p x + q=0 to find p= -3 and q= -1

now we solve
w^6 + q w^3 - p^3/27 =0
with our numbers
w^6 - 3 w^3 +1/27 = 0

let u= w^3, and solve
u^2 -3u + 1/27 = 0

Answer 8

would i simplify the 27?

Answer 9

\[3^{3}\]

Answer 10

wait. I got my p's and q's mixed up (I did not "mind by p's and q's"

Answer 11

now we solve
w^6 + q w^3 - p^3/27 =0
p= -3 and q= -1

so we should be doing
w^6 - w^3 - (-3)^3/27 = 0
w^6 - w^3 + 1 = 0

Answer 12

let u= w^3

u^2 - u + 1 = 0

Answer 13

ok im behind a bit where did the P,Q,and W comes from

Answer 14

I am (trying to ) doing this
https://en.wikipedia.org/wiki/Cubic_equation#Vieta.27s_substitution

Answer 15

ahh ok

Answer 16

\[u ^{2}-u=-1\]

Answer 17

do i divide the -u?

Answer 18

\[u ^{2}=\frac{ 1 }{ u }\]

Answer 19

then sqrt ?\[u=\sqrt{\frac{ 1 }{ u }}\]

Answer 20

either do the quadratic formula or complete the square.

Answer 21

the Quadratic formula is \[x=\frac{ -b \sqrt{b ^{2}-4ac} }{ 2a }\] correct with A=1 B=-1 and c=1?

Answer 22

\[x=\frac{ 1\sqrt{-1^{2}-4(1)(1)} }{ 2(1) }\]

Answer 23

\[x=\frac{ 1\sqrt{-3} }{ 2 }\]

Answer 24

\[x=\pm \frac{ -1\sqrt{-3} }{ 2 }\]

Answer 25

oops
\[x=\pm \sqrt{\frac{ -1\sqrt{-3} }{ 2 }}\]

Answer 26

you got confused.
it is (-b ± sqr(-3) )/2

or
(1 ± sqr(-3))/2

Answer 27

so
\[x=\frac{1\pm \sqrt{3} }{ 2 }\]

Answer 28

didnt think the other one looked right

Answer 29

u^2 - u + 1 = 0

a=1, b= -1, c=1

yes
\[ x=\frac{1\pm \sqrt{3} }{ 2 } \]

we now need the cube root of this ( gack!)

Answer 30

oh, the sqr(-3) is sqr(3) * i where i = sqr(-1)

Answer 31

we got a complex number.

we may be wandering into stuff you don't know?

in polar form,
\[ x = e^{i \frac{\pi}{3}} \]
the cube root is
\[ x^{\frac{1}{3}} = e^{i \frac{\pi}{3}\cdot \frac{1}{3}} =e^{i \frac{\pi}{9}} \]

Answer 32

according to https://en.wikipedia.org/wiki/Cubic_equation#Vieta.27s_substitution
one root is
t1= x - p/(3x)

p is -3 and x is e^(i pi / 9)
and we get
t1= e^(i pi/9) + 3/3e^(i pi / 9) = e^(i pi/9) + e^(-i pi/9) = 2 cos( pi/9)

Answer 33

-0.2repeating

Answer 34

?
one root is
2 cos(20º) = 1.87938524

the other two roots are equally spaced 120º apart
2 cos(120+20) = 2 cos(140)= 1.5321
2 cos(240+20) = 2 cos(260) = -0.3473

Answer 35

oh i just typed 2 cos pi over 9 into google

Answer 36

oh. you should put in parens
2*cos(pi/9)

because it did 2*cos(pi) / 9 which is different

Answer 37

1.87938524157

Answer 38

in radians the other two roots are
2 cos( 2pi/3 + pi/9) = 2 cos(7 pi/9)= -1.5321 (I left a minus sign out up above )
and
2 cos (4pi/3 + pi/9) = 2 cos(13 pi/9) = -0.3472963

Answer 39

im confused now whats all this?

Answer 40

the cubic has 3 solutions

Answer 41

ohh i forgot about that

Answer 42

according to https://en.wikipedia.org/wiki/Cubic_equation#Vieta.27s_substitution
the roots are
t= x - p/(3x)
with p= -3 we get
t= x + 1/x or
t= x + x^(-1)
when x is exp(i theta) this turns into 2 cos(theta)
t= 2 cos(theta)

where x is the cube root of
\[ u = \frac{1\pm \sqrt{3}i }{ 2 } \]
I found one cube root of u as e^( i*pi/9)
the other two cube roots are separated by 120º (2 pi/ 3 radians)
in other words we have
x= exp( i pi/9)
x= exp( i 7pi/9)
x= exp(i 13 pi/9)

Answer 43

these seems awfully complicated stuff.

Answer 44

very complicated never done anything like this