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anonymous
 one year ago
3x^29x12=0 by completing the square
with explanation if possible.
anonymous
 one year ago
3x^29x12=0 by completing the square with explanation if possible.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0divide the whole thing by 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[3(x^23x4) = 0 \rightarrow 3(x^23x)=4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to complete the square...you take the middle term and divide by 2 and square it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[3(x^23x+(\frac{ 3 }{ 2 })^2)=4+(\frac{ 3 }{ 2 })^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks like i forgot to x3 on the other end

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[3(x^23x+(\frac{ 3 }{ 2 })^2=12+3(\frac{ 3 }{ 2 })^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and also completing the square method

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you should get \[3(x^23x+(\frac{ 9 }{ 4 })=18\frac{ 3 }{ 4 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you can factorize the\[3(x^23x+\frac{ 9 }{ 4 })\]part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, thank you very much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can factorize that on your own?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, i was having trouble getting to that point though. appreciate your help!

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0Start with your equation: \(3x^29x12=0 \) Notice that there is a common factor of 3 for all terms, so divide both sides by 3: \(x^23x4=0 \) Now add 4 to both sides. \(x^23x=4 \) The term that needs to be added to the left sides is obtained this way. Take half of the xterm coefficient and square it. Half of 3 is \(\dfrac{3}{2}\). The square of that is \(\dfrac{9}{4} \) Now we add \(\dfrac{9}{4} \) to both sides \(x^23x + \dfrac{9}{4}=4 + \dfrac{9}{4}\) We rewrite the left sides as the square of a binomial and we add the numbers on the right side. \(\left( x  \dfrac{3}{2} \right)^2 = \dfrac{16}{4} + \dfrac{9}{4} \) \(\left( x  \dfrac{3}{2} \right)^2 = \dfrac{25}{4} \) Now we apply the rule: If \(x^2 = k\), then \(x = \pm \sqrt k\) to get: \( x  \dfrac{3}{2} = \pm \sqrt{\dfrac{25}{4}} \) \( x  \dfrac{3}{2} = \pm \dfrac{5}{2} \) We separate the equation into two parts: \( x  \dfrac{3}{2} = \dfrac{5}{2} \) or \( x  \dfrac{3}{2} = \dfrac{5}{2} \) Add \(\dfrac{3}{2} \) to both sides in both equations: \( x = \dfrac{8}{2} \) or \( x = \dfrac{2}{2} \) Reduce the fractions: \( x = 4 \) or \( x = 1 \)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0The step of completing the square by adding the square of half of the xterm coefficient to both sides only works if the coefficient of the x^2 term is 1. After we divided the equation by 3 on both sides, we did have a coefficient of 1 for the x^2 term, so we were able to complete the square that way.

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0i agree with you

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0i had previously solved the sum in that way but i deletedit
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