Solving Equation: x^5 + k^2x = 0

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The equation x^5 + k^2x = 0 has multiple solutions, including x = 0. Dividing by x is only valid if x is not zero, which is a crucial point in solving the equation. The correct approach involves factoring the equation rather than prematurely dividing. The discussion highlights the importance of checking conditions before performing algebraic manipulations. Understanding these concepts is essential for correctly solving polynomial equations.
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Homework Statement



I am asked to solve the following equation, giving answer in terms of k

Homework Equations



$$x^5 + k^2x = 0$$

The Attempt at a Solution



The answer is apparently 0. What is 0. Not even sure what that means.

I would have thought: divide through by x to obtain

$$k^2 = -x^4$$ →

$$ k=x^2i $$ ?
 
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You solved for k. I think you were supposed to solve for x.

I agree with you that there is more than one root.
 
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Atomised said:

Homework Statement



I am asked to solve the following equation, giving answer in terms of k

Homework Equations



$$x^5 + k^2x = 0$$

The Attempt at a Solution



The answer is apparently 0. What is 0. Not even sure what that means.

I would have thought: divide through by x to obtain

$$k^2 = -x^4$$ →

$$ k=x^2i $$ ?
Dividing both sides of an equation by anything except zero, gives you an equivalent equation.

You can divide by x, as long as it's not equal to zero.
What if x is zero?

Evaluate that case another way, for instance, by plugging zero in for x.​
 
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Atomised said:

Homework Statement



I am asked to solve the following equation, giving answer in terms of k

Homework Equations



$$x^5 + k^2x = 0$$

The Attempt at a Solution



The answer is apparently 0. What is 0. Not even sure what that means.
The "answer" is not zero. Your answer should be in the form of equations that start with "x = ..."

One of the solutions is x = 0, but there is another. Factoring the left side would be helpful.
 
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Atomised said:

Homework Statement



I am asked to solve the following equation, giving answer in terms of k

Homework Equations



$$x^5 + k^2x = 0$$

The Attempt at a Solution



The answer is apparently 0. What is 0. Not even sure what that means.

I would have thought: divide through by x to obtain

$$k^2 = -x^4$$ →

$$ k=x^2i $$ ?

You have committed the worst sin in mathematics, viz., dividing by x before checking that it is allowed. If x = 0 you cannot do any such division---but in that case, you don't need to anyway. If x ≠ 0 then---and only then---can you divide both sides by x.
 
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Thanks Mark & Ray for these lessons - a great help in learning to think properly.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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