Question

can someone please show me how my teacher got an answer. I can't figure it and I need some help.

Answer 1

\[w ^{4}-14w ^{2}-2=0\]

Answer 2

\[Let: u =w ^{2}\]

Answer 3

\[u ^{2}+14u-2=0\]

Answer 4

\[u=7-\sqrt{51}\]

Answer 5

Quadratic formula again!

Answer 6

or completing the square.

Answer 7

i need to know how my teach got the result 51

Answer 8

or graph it!

Answer 9

so nobody knows. that's it, huh

Answer 10

that's all you had to say.

Answer 11

Aww Notebook, you teacher could have any of the above methods I mentioned.

Answer 12

Just relax and think.

Answer 13

substituting w= 51 in the eq. above you cannot get 0

Answer 14

It doesn't the polynomial have no rational solution.

Answer 15

roots*

Answer 16

the user is no more on this question lol

Answer 17

you are on the right track: \[w = \pm \sqrt{7\pm \sqrt{51}}\]

Answer 18

The quadratic formula for
ax^2+bx+c=0
gives
x=(-b+/-sqrt(b^2-4ac))/2a
substitute
a=1
b=14
c=-2
and x=u
you'll get
u=(-14+/-sqrt(14^2+8))/2
=(-14+/-sqrt(204)/2
=-7 +sqrt(51) or -7-sqrt(51)

Answer 19

@Pax are you sure it's \( w = \pm \sqrt{7\pm \sqrt{51}} ) not \( u = \pm \sqrt{7\pm \sqrt{51}} \) ? ;)

Answer 20

Oh, sorry, the original equation was
u^2-14u-2=0
so the solution should read:
u=7 +sqrt(51) or 7-sqrt(51)

Answer 21

\[u=w ^{2}=7\pm \sqrt{51}\] so \[w=\pm \sqrt{7\pm \sqrt{51}}\]
if w has to be real you would remove \[w=\pm \sqrt{7- \sqrt{51}}\]