anonymous
  • anonymous
3x^2-9x-12=0 by completing the square with explanation if possible.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
divide the whole thing by 3
anonymous
  • anonymous
x^2-3x-4=0
anonymous
  • anonymous
\[3(x^2-3x-4) = 0 \rightarrow 3(x^2-3x)=4\]

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anonymous
  • anonymous
to complete the square...you take the middle term and divide by 2 and square it
anonymous
  • anonymous
\[3(x^2-3x+(\frac{ 3 }{ 2 })^2)=4+(\frac{ 3 }{ 2 })^2\]
anonymous
  • anonymous
looks like i forgot to x3 on the other end
anonymous
  • anonymous
\[3(x^2-3x+(\frac{ 3 }{ 2 })^2=12+3(\frac{ 3 }{ 2 })^2\]
anonymous
  • anonymous
its 3 not 4
anonymous
  • anonymous
and also completing the square method
anonymous
  • anonymous
you should get \[3(x^2-3x+(\frac{ 9 }{ 4 })=18\frac{ 3 }{ 4 }\]
anonymous
  • anonymous
now you can factorize the\[3(x^2-3x+\frac{ 9 }{ 4 })\]part
anonymous
  • anonymous
you catching on?
anonymous
  • anonymous
yes, thank you very much
anonymous
  • anonymous
you can factorize that on your own?
anonymous
  • anonymous
yes, i was having trouble getting to that point though. appreciate your help!
anonymous
  • anonymous
yw
mathstudent55
  • mathstudent55
Start with your equation: \(3x^2-9x-12=0 \) Notice that there is a common factor of 3 for all terms, so divide both sides by 3: \(x^2-3x-4=0 \) Now add 4 to both sides. \(x^2-3x=4 \) The term that needs to be added to the left sides is obtained this way. Take half of the x-term 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 \(x^2-3x + \dfrac{9}{4}=4 + \dfrac{9}{4}\) 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
  • mathstudent55
The step of completing the square by adding the square of half of the x-term 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
  • madhu.mukherjee.946
i agree with you
madhu.mukherjee.946
  • madhu.mukherjee.946
i had previously solved the sum in that way but i deletedit

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