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12132006  #1 
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Can somebody please help me
I have a test tomorrow. First if any one knows any good Math A sites? Plus would anyone know how to solve this problem? Could you explain it? Please.
Solve the following quadratic equation. x23x10=0 
12132006  #2 
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I dont know, but I boiled it down to x.x10=3x

12132006  #3 
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............................2
I found the answer. x3x10=0. (The 2 is an power for x. Forget the Dots) It is (x5)(x+2) 5,2 
12132006  #4 
Join Date: Oct 2006
Posts: 822

the answer is x=5, or 2
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12132006  #5 
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Can you explain to me how you came up with that solution?

12132006  #6 
Join Date: Oct 2006
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x^23x10=0
that's how it starts (this is a quadratic equation) then i factored for exampla: (xm) (x+n) and the original is ax^2+bx+c=0 m+n must equal b, and mn must equal c. 5 and 2 add up to 3 and they multiply to get 10 therefore, it is (x5)(x+2)=0 either x5=0, or x+2=0 therefore x=5, or 2 now who do u think is going to win mathcounts? do u get how i did it?
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