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Mathematics 18 Online
OpenStudy (mb194227):

Please help me solve x^3-6x^2+6x=0.

OpenStudy (3mar):

Could I help you?

OpenStudy (kira_yamato):

Hint: \[x = \frac{ -b ± \sqrt{b^2 - 4ac} }{ 2a }\]

OpenStudy (3mar):

Hey, @Kira_Yamato! It is a cubic equation, not quadratic!

OpenStudy (mb194227):

Could I perhaps split the middle term so that I can solve the equation by grouping? I'm open to any other suggestions.

OpenStudy (3mar):

I don't think so. But what I am sure about is that you can take xa s a common factors from all three terms! agree with me?

OpenStudy (mb194227):

Yes, so I get x(x^2-6x+6)=0. What now?

OpenStudy (3mar):

You are Awesome!

OpenStudy (3mar):

Now you have a quadratic equation which could be solved by the quadratic formula! Are you familiar with it?

OpenStudy (mb194227):

I used the quadratic formula to find that x=3+sqrt(3) and x=3-sqrt(3), so can x=0 be a solution?

OpenStudy (kira_yamato):

Technically I'll say factorization of x yields a quadratic equation

OpenStudy (phi):

yes, when you write x (x^2-6x+6)=0 that is like A*B=0 if A is 0 you will get 0 if B is 0 you get 0 in this case if x=0 you get 0, so x=0 is a solution. and of course of x^2-6x+6=0 gives two more solutions

OpenStudy (3mar):

"so can x=0 be a solution?" Yes, Of course, in addition to the two solutions you have gotten from the qu.formula!

OpenStudy (mb194227):

Thank you all for helping me solve this equation!

OpenStudy (3mar):

That is with my pleasure! Any Help... Any Time... Are you persuaded?

OpenStudy (mb194227):

Of course!

OpenStudy (3mar):

I am happy to hear that. Thank you for learning!

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