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How do you solve #x^5-5x^3+4x=0#?

Answer 1

Kennedy Adams

#x=0#, #x=1#, #x=2#, #x=-1#, and #x=-2#

Start off by factoring out an #x# as follows

#x(x^4-5x^2+4)=0#

We can factor #x^4-5x^2+4#

Doing so we have

#x(x^2-4)(x^2-1)=0#

So in order for this to be equal to zero

#x=0#

#x^2-4=0#

#x^2-1=0#

Well #x=0#

#x^2=4#

#x=+-sqrt(4)=+-2#

#x^2=1#

#x=+-sqrt(1)=+-1#

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Answer 2

Kennedy Adams

To solve the equation (x^5 - 5x^3 + 4x = 0), you can factor out (x), giving you (x(x^4 - 5x^2 + 4) = 0). Then, you solve for (x) by setting each factor equal to zero:

(x = 0)
(x^4 - 5x^2 + 4 = 0)

For the second equation, you can substitute (y = x^2), giving you (y^2 - 5y + 4 = 0). Then solve for (y) using the quadratic formula:

[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}]

where (a = 1), (b = -5), and (c = 4).

Solve for (y), then find (x) by taking the square root of the solutions for (y). Keep in mind that when taking the square root, you'll get two solutions for each (y) solution.

Once you have all solutions for (x), which include (x = 0) and the solutions from the quadratic equation, those are the solutions to the original equation.

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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.