Question

help please (:

find the zeros of the function. state the multiplicity of any multiple zeros.

17. y=x^4 - 8x^2+ 16

Answer 1

first zero is x=0, then y = 16
second zeros you get by changing x^2 with t, then you get
t^2 - 8t +16 = 0, and the rest is quadratic equation :)

Answer 2

you get t1 and t2, and change both in x^2 = t and get your answers for y=0 :)

Answer 3

do you get it or do you need more help?

Answer 4

more help..

Answer 5

okay, quadratic equation goes ax^2+bx+c =0
so in this case, a = 1, b=-8, c=16
next formulas are:
\[t1= (-b+\sqrt{b ^{2}-4*a*c})/(2*a)\] and \[t1= (-b-\sqrt{b ^{2}-4*a*c})/(2*a)\]

Answer 6

*second one is t2

Answer 7

but in this case, both of them will give you the answer 4, so it makes job easier
\[t = x ^{2}, x=\sqrt{t}, x =\sqrt{4}\]

Answer 8

and you have two solutions here (-2,0) and (2,0), since both (-2)^2 and 2^2 are 4, and also from beginning (0,16)

Answer 9

and that's all there is :) hope it's all clear now

Answer 10

ok i think i kind of get it now thankss (:

Answer 11

no problem :)