Question

4w^4+40w^2-44=0

Answer 1

divide both sides by 4 first!

Answer 2

\[w^4+10w^2-11=0\]

Answer 3

now let u=w^2

Answer 4

\[u^2+10u-11=0\]

Answer 5

that should be easier to solve now

Answer 6

please type a very elaborate reply on how you find the real number solutions

Answer 7

Divide throughout by 4.

\[w^4 + 10 w^2 - 11 = 0\]\[(w^2 + 11)(w^2 - 1)\]

Answer 8

=0

Answer 9

\[ 4 (w-1) (w+1) \left(w^2+11\right) \]

Answer 10

w2 =y
4y2 +40y -44=0 /4
y2 +10y -11=0

y_1,_2=(-10+/- sqrt(100+44))/2 =(-10+/- 12)/2 = -22/2 = -11 and 2/2 =1

w2= -11
w_1,2= +/- isqrt11

w2=1
w_3,4= +/- 1

Answer 11

so the solutions are \(1,-1\sqrt{11} i \)

Answer 12

w here will be real

Answer 13

*\( 1,-1,\sqrt{11} i \)

Answer 14

lol

Answer 15

oh sorry
we get w^2=-11 or 1
w=+-1 or +-(11i)^1/2

Answer 16

yes the product of plus or minus the square root of 11 and i

Answer 17

4 roots in all

Answer 18

yep