SOLUTION: how would you solve the problem: m^16-25=0 and set it equal to zero... i don't know how to do it with m^16. but a normal one would look like: 16y^2-25=0 (4y-5)(4y+5)=0

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Question 32920: how would you solve the problem:
m^16-25=0
and set it equal to zero... i don't know how to do it with m^16. but a normal one would look like:
16y^2-25=0
(4y-5)(4y+5)=0
4y-5=0 4y+5=0
4y=5 4y=-5
y= 5/4 y= -5/4
but i dont know how to do it with m^16...
i get as far as:
m^16-25=0
(m^8-5) (m^8+5)=0
m^8-5=0 m^8+5=0
m^8= 5 m^8= -5
please do it and show the work...thanks :)

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
You are doing fine when you wrote:
(m^8-5)(m^8+5)=0
Therefore m^8=5 or m^8=-5
m=5^(1/8)= 1.223..
You could also say m^8+5=0
So, m^8=-5. This is true only
if m is a complex number.
I'm not saying the above is the only
way you could solve m^16-25=0; it depends
on what level of mathematics you are studying.
But your writing (m^8-5)(m^8+5)=0 is the way
to start.
Cheers,
Stan H.

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
You did all the work. It looks OK to me
All you need to add is the las steps
(m^8-5)
(m^8+5)=0
m^8-5=0
m^8+5=0
m^8= 5
m^8= -5


check by substitution





and




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