How do you solve #7x ^ { 3} + 21x ^ { 4} = 0#?

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Answer 1 · 2018-03-09T03:25:51

See a solution process below:

First, factor out the common term #(7x^3)# from each term on the left side of the equation:

#(7x^3 xx 1) + (7x^3 xx 3x) = 0#

#7x^3(1 + 3x) = 0#

Now, solve each term on the left for #0#:

Solution 1:

#7x^3 = 0#

#(7x^3)/color(red)(7) = 0/color(red)(7)#

#x^3 = 0#

#root(3)(x^3) = root(3)(0)#

#x = 0#

Solution 2:

#1 + 3x = 0#

#1 - color(red)(1) + 3x = 0 - color(red)(1)#

#0 + 3x = -1#

#3x = -1#

#(3x)/color(red)(3) = -1/color(red)(3)#

#x = -1/3#

The Solution Set Is:

#x = {-1/3, 0}#