Question 4587: I need help with these two problems.
x^4 + 5x^3 + x^2 - 25 =0 and 3x^(3/2) + 7x^(1/3) = 6.
I tried to group the first one but got stuck. I have no idea how to do the second problem. Need help before Monday
Nicole'
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! (i) x^4 + 5x^3 + x^2 - 25 =0
x^3(x + 5) + x^2 - 25 =0
x^3(x + 5) + (x+5)(x-5) =0
(x+5)(x^3 + x - 5) = 0
So, x = -5 or x^3 + x - 5= 0
However,unfortunately,it is impossible to factor out
x^3 + x - 5 =0 unless you use the Cartan's formula about the cubic
polynomial to find the roots.
If you go to quickmath.com you can get the complicated
solutions of 1 real and two conjugate complex numbers.
But I think it is beyond what your level of learning.
[Maybe you only need one root.]
(ii) 3x^(3/2) + 7x^(1/3) - 6 = 0.
Let u = x^(1/3), we have
3 u^2 + 7 u - 6 = 0.
Factor out: (3u + 2)(u -3) =0
So, u = -2/3 or 3.
Hence, x^(1/3) = -2/3 , x = -(2/3)^3 = -8/27
or x^(1/3) = 3 , x = 27.
Plz check by yourself.
Kenny
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