Question 1013108: Solve x^4 +4x^2 = 32. Find 4 roots.
Thanks. Found 4 solutions by fractalier, ikleyn, macston, josmiceli:Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website! Just did this for you...
Okay, from
x^4 + 4x^2 = 32
x^4 + 4x^2 - 32 = 0
we can factor and get
(x^2 + 8)(x^2 - 4) = 0
(x + 2isqrt(2))(x - 2isqrt(2))(x + 2)(x - 2) = 0
giving us four roots
x = +/- 2isqrt(2) and x = +/- 2 Answer by ikleyn(52879) (Show Source): You can put this solution on YOUR website! .
Solve x^4 +4x^2 = 32. Find 4 roots.
-----------------------------------------
= 32,
- 4 = 32,
= 36,
= +/- 6.
Now solve two quadratic equations
1. = 6 ----> = 4 ----> Two solutions: x = 2 and x = -2.
2. = -6 ----> = -8 ----> Two solutions: x = and x = .
You obtain four solutions, in total.
Answer by macston(5194) (Show Source): You can put this solution on YOUR website! .
Let y=x^2
.
x^4+4x^2=32
x^4+4x^2-32=0 . Substitute y for x^2
y^2+4y-32=0
(y+8)(y-4)=0
y+8=0 --OR-- y-4=0
y=-8 --OR-- y=4 . Replace y with x^2
x^2=-8 --or-- x^2=4
x=+/- --OR-- x=+/-2
x=+/- --OR-- x=+/-2
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website! Let
Substitute in the equation
Complete the square:
Take the square root of both sides
-----------------
-------------
and
-------------------
The 4 roots are:
---------------------
You can plug these back into the
equation to check