SOLUTION: how would you solve 16h to the 5th minus 25 h to the 3rd plus 9h equals 0

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Question 812219: how would you solve 16h to the 5th minus 25 h to the 3rd plus 9h equals 0
Found 2 solutions by Edwin McCravy, KMST:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
16h5-25h3+9h = 0

Factor out common factor h

h(16h4-25h2+9) = 0

Factor the trinomial in the parentheses

h(16h2-9)(h2-1) = 0

Each of those binomial in parentheses are
the difference of squares, so factor them:

h(4h-3)(4h+3)(h-1)(h+1) = 0

By the zero-factor property set all those factors = 0:

h=0;  4h-3=0;  4h+3=0 ;   h-1=0;  h+1=0
        4h=3     4h=-3      h=1     h=-1
         h=3%2F4      h=-3%2F4

So the solutions are 0, 3%2F4, -3%2F4, 1, and -1

Edwin

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
16h%5E5-25h%5E3%2B9h=0
%2816h%5E4-25h%5E2%2B9%29h=0 --> highlight%28x=0%29 is one solution
The other solutions, if there are any, must come from
16h%5E4-25h%5E2%2B9=0
I know that I can factor 16y%5E2-25y%2B9=%2816y-9%29%28y-1%29 , so
16h%5E4-25h%5E2%2B9=%2816h%5E2-9%29%28h%5E2-1%29
and that can be applied to transform the equation:
16h%5E4-25h%5E2%2B9=0
%2816h%5E2-9%29%28h%5E2-1%29=0
%284h%2B3%29%284h-3%29%28h%2B1%29%28h-1%29=0
So, the other solutions are
highlight%28h=-3%2F4%29 , highlight%28h=3%2F4%29 ,
highlight%28h=-1%29 , and highlight%28h=1%29 .

NOTE: You could also say that from 16h%5E4-25h%5E2%2B9=0 ,
we do the change of variable y=h%5E2
to get the quadratic equation 16y%5E2-25y%2B9=0 ,
with solutions y=1 and y=9%2F16
and returning from y to h we get
h%5E2=1 --> system%28h=1%2C+%22or%22+%2Ch=-1%29
and
h%5E2=9%2F16 --> system%28h=3%2F4%2C+%22or%22+%2Ch=-3%2F4%29 .