Question

x^4+13x^2+36=0

Answer 1

@Neemo

Answer 2

:) let c=x^2 and find an equation satisfied by c !

Answer 3

what would be ?!

Answer 4

idk

Answer 5

ok ! if c=x^2 ! the x^4=??

Answer 6

then ** x^4=(something with c ...just c) :)

Answer 7

where are u getting c from

Answer 8

im srry i dont understand

Answer 9

"we" don't know how to solve equation equation that contain x^4, yet...just a trick ! that you should learn :) !

Answer 10

ok

Answer 11

so c is a new unkown number ! c=x^2
so, if c=x^2 then x^4= ???(remember with jyst c)

Answer 12

:/ c

Answer 13

ok you did'nt like that "c" don't you ?
I'll make it easier for you ! if c = x^2 ...\[c=x^2\]

you may know that \[x^4=(x^2)^2\] do you know that ?!

Answer 14

@annej It's easy when you know how it works ! Trust me !

Answer 15

still there ! ?! @annej

Answer 16

srry yes

Answer 17

ok whats next

Answer 18

:) ! nno problem !

Answer 19

ok sooo what do i do now

Answer 20

c= x^2

Answer 21

yeah ! so x^4 = ?!

Answer 22

x^2 :D

Answer 23

c=x^2 ====>x^4=....with c :) !

Answer 24

:/ idk

Answer 25

it's c^2 :)

Answer 26

look \[x^2=c=====>x^4=(x^2)^2=(c)^2\]

Answer 27

im sure your getting very frustrated with me but i dont see what this has to do with the above problem

Answer 28

do i use the quadratic formula , i dont think thats possible with this problem , do i break it down

Answer 29

kk we will do it !
look c=x^2 then c^2=x^4
so the equation satisfied by c is \[c^2+13c+36=0 \]
now you can use the quadratic (formula with the unknown c) !

Answer 30

ok i understand, that , my problem is the c where did it come from and how did u get rid of x^4

Answer 31

y does c= x^2 is that jut how it is

Answer 32

c is just a new unkown ! you can choose another name of "c" which one you like y :) !
and the equation tells me that look
\[(x^2)^2+13(x^2)+36=0\] now you see ! why I took x^2=aor b or c...or y or z....

Answer 33

so \[(x^2)\] should be the unknown to apply quadratic formula

Answer 34

ohhh ok gocha srry for my slowness

Answer 35

no no don't worry ! now we didn't finish yet !

Answer 36

so can you find the "c" ! :)

Answer 37

i can do it from here

Answer 38

yeah you know the equation ! c^2+13c+36=0

Answer 39

or in other words ! can you find \[(x^2)\]

Answer 40

is it -13pm5/2

Answer 41

I don't know ! :) did you use your calculator cause I don't have one ! Sure...:) ?!

Answer 42

nvm its -4 and -9

Answer 43

sure....?! kk I'll check!

Answer 44

13^2-4*36=25
so you're right !

Answer 45

now ! c=(x^2)=-4 or c=(x^2)=-9
is that possible ?!

Answer 46

please don't leave me alone again :'( !

Answer 47

srry umm no

Answer 48

It's not possible ! are you sure about that ? :)
(If you want I can give you all tricks that can exist about quadratic formula)

Answer 49

ok

Answer 50

now ! Tell me why x^2=-4 or x^2=-9 is not possible !

Answer 51

:(

Answer 52

x^2 is always positive...so it can't be equal to a negative number see ?!

Answer 53

yes

Answer 54

so my answer is in correct

Answer 55

no !who said that ! so the set of solutions=

Answer 56

no !who said that ! so the set of solutions=\[\emptyset \]

Answer 57

is empty ! there is no real solutions to that equation !

Answer 58

sorry !! I have to go now :'(! try to solve these two equations using the same idea:
\[2x-\sqrt{x}-1=0\]
\[3x^2+2\left| x \right|-1=0\]
just choose the new unknown okk .?