Question

x^4 + 3x^2 - 2 = 0 Solve for x

Answer 1

Let a=x^2 then
x^4 + 3x^2 - 2 = 0
(x^2)^2+3x^2-2=0
a^2+3a-2=0

Answer 2

Let \(x^2 = y\):\[y^2 + 3y - 2 = 0 \]Solve for \(y\).

Answer 3

Now, can u solve for a

Answer 4

can i use z next?

Answer 5

@lgbasallote Be my guest ;)

Answer 6

I know about solving this with substitution. I just got a weird number so I'm just wondering it that's right

Answer 7

let \(z = x^2\)
\[\implies z ^2 + 3z - 2 = 0\]
factor out to solve for z

Answer 8

and no..it shouldn't be a weird number

Answer 9

on second thought..it is

Answer 10

you have to use Q.F. and not factoring

Answer 11

USe quadratic formula

Answer 12

\[z = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \]

Answer 13

you need to use quadratic formula and the result is weird number....

Answer 14

I hate weird @Numbers.

Answer 15

her engagement got a lot down since i dethroned her....

Answer 16

agdg's too....

Answer 17

@rebecca1233 now has the second highest engagement

Answer 18

wonder when she'll dethrone me >:))

Answer 19

@lgbasallote funny haan!!!!

Answer 20

Weird thing is that agd's SmartScore is still increasing.

Answer 21

Satellite will be dethroned by @lgbasallote

Answer 22

i have to go through the triumvirate first =_= the royal bodyguards of zeus

Answer 23

\[\sqrt{\frac{ \pm \sqrt{17}-3 }{ 2 }}\]

Answer 24

I need to stop underestimating myself. Got it right first time

Answer 25

...wait..why is there a big square root?

Answer 26

@ArkGoLucky thats bad.....

Answer 27

@lgbasallote first you z came than the y... lolzzzZZZ...

Answer 28

oh sorry then x...

Answer 29

i didn't want to look as if i copied @ParthKohli

Answer 30

anyway...what's with the big square root?