Constant determining a double root

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SUMMARY

The function y=x^3 - 2x^2 + k has a double root when k=0, as this allows for the factorization (x^2)(x-2)=0, resulting in a bounce at x=2. A double root occurs when both the function and its derivative equal zero at the same point. To find another value of k that ensures a double root, one must solve the derivative f'(x) = 3x^2 - 4x and set it equal to zero, identifying potential double root locations and selecting k accordingly.

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Consider the function y=x^3 - 2x^2 + k, where k is a constant. Explain why k=0 ensures that f(x)=0 has a double root. A double root is a bounce and I thought that when k=0, the x^2 can be immediately common factored out of the function, so you have (x^2)(x-2)=0. Therefore, you will always have a bounce with that x^2. Am I even close?

Also, I have to determine another value of k that ensures f(x)=0 has a double root. This I am completely stuck on...
 
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If f(x) has a double root at x=a, then f(a)=0 and the derivative f'(a)=0. Yes, this is because it has a 'bounce'. Sort of. It doesn't bounce in this case. But it does have zero derivative. Solve the derivative to find where double roots can be and then pick k to make them also roots.
 

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