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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Radicals
-> 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.
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
(