Question

find the real
roots of the equaton 18/x^4+1/x^2=4

Answer 1

The Real roots of the given equation is X = 3/2 and -3/2

Multiply with x^4 in the whole equation
Then there will be a equation = 4x^4 - x^2 - 18 = 0
On solving, u will get equation = ( x^2 + 2)(4x^2 - 9) = 0
Solve for the real part & and u will get the Answer..!!

Answer 2

how did u get ( x^2 + 2)(4x^2 - 9

Answer 3

@Trexy

Answer 4

wait.. XD

Answer 5

Use ac test to factor. First, you factor 4 into 4 and 1, -18 into -9 and 2 as below
4 -9
1 2
Then you cross multiply to make sure you have -1.
Then add x to the right column to write
(4x^2 - 9)(x^2 +2)

Answer 6

Sorry about that.. anyway.. he got the answer...

Answer 7

how do u factor out 4 into four into1?um a bit puzzled

Answer 8

Hahah.. ok wait.

Answer 9

@Trexy

Answer 10

@Mr._To

Answer 11

because 4*1=4 which gives your x^4 coefficient. The 2 and -9 give you back your -18 constant. Now all you need is the coefficient for x^2. you're trying to determine if the 4 and 1, -9 and 2, factorization will give you back your -1 (for the x^2) upon expansion. It does. So you end up with the 4 and 1 attached to your x^2 terms and 2 and -9.
(4x^2-9)(x^2+2)=0

Answer 12

@Tushara excellent solution!

Answer 13

:) i hope that helped

Answer 14

Thanks @Tushara and @eseidl helped a lot :)

Answer 15

m glad, np