Why do you solve this? I know you have to use long division and remainder theorem...and something like that?

Question

2x^6 - 24x^3 + 4 = 0

Please assist!!

Answer 1

let x^3=u

2u^2-24u+4=0

u=[24+-sqrt(24^2-4*4*2)]/4

u=6+-sqrt(34)

x^3=6+√34 or x^3=6-√34

x=cubic root of 6+-√34

Answer 2

s777s347

let x^3=u

2u^2-24u+4=0

u=[24+-sqrt(24^2-4*4*2)]/4

u=6+-sqrt(34)

x^3=6+√34 or x^3=6-√34

x=cubic root of 6+-√34






Chanche x^3 into y
2y^2 - 24y +4 =0
Solve y
Find x


V V

ok in this you let x^6=X^2
which therefore makes X^3=x
and then solve it quadratically until you come to this step

x = 6+root34 or x=6-root34
and then substitute for x=x^3 and solve
i'm pretty sure this is how you do it...hope that helps








let x^3 be t
then the given equation will reduce to 2t^2-24t+4=0
now taking out 2 as a common factor 2(t^2-12t+2)=o
=>t^2-12t+2=0
now applying the quadratic formula
t =[-(-12)+(144-8)^1/2/2or [-(-12)-(144-8)^1/2]/2
the formula being the solution of ax^2+bx+c=0 is given by
-b plus or minus (b^2-4ac)whole divided by 2a
now whatever value tou get for t will be the value of x^3
thus the value of x will be the cube root of t




Why do we solve this? Because it's there!




On this question, I would say Hm-mm.....

Well, I can divide by 2:

x^6-12x^3+2=0

It's only got three terms, too bad it's not a quadratic!


I'd beat my head against my desk a few time, my eyes would cross...
But ya know, the exponent on the first term is the SQUARE of that on the second term and this thing is looking even MORE like a quadratic....

Can I MAKE it one? What would that look like?

(x^3)^2-12(x^3)+2=0

So,I do have a quadratic, if my variable is x^3.

I apply the quadratic formula, which is tattooed over my heart. (It IS tattooed over YOUR heart, RIGHT?)

DON'T LOOK NOW, YOU'RE IN PUBLIC! (Yahoo Answers:)

Take the cubed root of those roots.
Say OMG what an UGLY mess! This can't possibly be right!

So I substitute in the ORIGINAL equation.

Viola! Success!

Source(s):

BA in Mathematics.

Answer 3

If 2x^6 - 24x^3 + 4 = 0
then 2(x^6 - 12x^3 +2) = 0
then (x^6 - 12x^3 + 2) = 0 because 2 x 0 = 0
If I could factor (x^6 - 12 X^3 + 2) I would know the two possible answers, but I don't know how to factor it! Sorry. I had fun trying though.

Answer 4

On this question, I would say Hm-mm.....

Well, I can divide by 2:

x^6-12x^3+2=0

It's only got three terms, too bad it's not a quadratic!


I'd beat my head against my desk a few time, my eyes would cross...
But ya know, the exponent on the first term is the SQUARE of that on the second term and this thing is looking even MORE like a quadratic....

Can I MAKE it one? What would that look like?

(x^3)^2-12(x^3)+2=0

So,I do have a quadratic, if my variable is x^3.

I apply the quadratic formula, which is tattooed over my heart. (It IS tattooed over YOUR heart, RIGHT?)

DON'T LOOK NOW, YOU'RE IN PUBLIC! (Yahoo Answers:)

Take the cubed root of those roots.
Say OMG what an UGLY mess! This can't possibly be right!

So I substitute in the ORIGINAL equation.

Viola! Success!

Answer 5

let x^3 be t
then the given equation will reduce to 2t^2-24t+4=0
now taking out 2 as a common factor 2(t^2-12t+2)=o
=>t^2-12t+2=0
now applying the quadratic formula
t =[-(-12)+(144-8)^1/2/2or [-(-12)-(144-8)^1/2]/2
the formula being the solution of ax^2+bx+c=0 is given by
-b plus or minus (b^2-4ac)whole divided by 2a
now whatever value tou get for t will be the value of x^3
thus the value of x will be the cube root of t

Answer 6

Chanche x^3 into y
2y^2 - 24y +4 =0
Solve y
Find x

Answer 7

ok in this you let x^6=X^2
which therefore makes X^3=x
and then solve it quadratically until you come to this step

x = 6+root34 or x=6-root34
and then substitute for x=x^3 and solve
i'm pretty sure this is how you do it...hope that helps

Answer 8

Why do we solve this? Because it's there!

Answer 9

my answer is same with the answer of
s777s347