Question

if the equation 2x^3 +kx - 4 =0 and 6x^4 + 3kx^2 + 2 =0 have a common root k equals ?

Answer 1

@Callisto @experimentX @badi

Answer 2

Can I do it in a very.... stupid way?
2x^3 +kx - 4 =0
kx = 4-2x^3

6x^4 + 3kx^2 + 2 =0
6x^4 + 3(kx)x + 2 =0
6x^4 + 3x(4-2x^3) +2 =0
6x^4 + 12x - 6x^4 +2 =0
Solve x

Answer 3

Then, sub. x into the equation and find k

Answer 4

@Callisto by ur method i got k=71 but the answer should bek=-433/18

Answer 5

Nope... I got your answer :|

Answer 6

I also got -433/18 for k.
x=-1/6 and if you put this into k=(4-2x^3)/x you get the k you want.

Answer 7

2x^3 +kx - 4 =0
kx = 4-2x^3

6x^4 + 3kx^2 + 2 =0
6x^4 + 3(kx)x + 2 =0
6x^4 + 3x(4-2x^3) +2 =0
6x^4 + 12x - 6x^4 +2 =0
12x = -2
x= -1/6

kx = 4-2x^3
k = (4-2x^3) / x
= (4-2(-1/6)^3) /(-1/6)
= -433/18

Answer 8

let common root be c
so we have
2c^3 + kc -4 =0 {multiply this by 3c}
=>6c^4 + 3kc^2 - 12c=0
and 2nd is 6c^4 + 3kc^2 + 2=0
=>c=-1/6
thus,on substituting,,
you should get your ans,,

Answer 9

@Callisto sorry that my calculation was wrong

Answer 10

Never mind :)

Answer 11

anywhere thanzz a lot....

Answer 12

x=-1/6
k=(4-2(-1/6)^3)/(-1/6)=-433/18

Answer 13

welcome!!~