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

Answer 1

Savannah Anderson

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#:

First Solution:

#7x^3 = 0#

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

#x^3 = 0#

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

#x = 0#

Option 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 Set of Solutions Is:

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

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Answer 2

Julia Adams

To solve the equation 7x^3 + 21x^4 = 0, you can factor out the common term x^3, leaving you with x^3(7 + 21x) = 0. Then, you set each factor equal to zero and solve for x. This gives you two solutions: x = 0 and 7 + 21x = 0, which simplifies to x = -1/3. So, the solutions to the equation are x = 0 and x = -1/3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.