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avai70178

  • one year ago

Could someone help me solve this equation by completing the square? 6a^2-12a-93=-3

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  1. Astrophysics
    • one year ago
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    To complete the square \[x^2+ax=0\] we can do the following \[x^2+\frac{ 2ax }{ 2 } + \left( \frac{ a }{ 2 } \right)^2-\left( \frac{ a }{ 2 } \right)^2=0\]

  2. Astrophysics
    • one year ago
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    \[6a^2-12a-90=0\] set it up as such now complete the square

  3. avai70178
    • one year ago
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    Wouldn't we add 93 to both sides so that we have \[6a ^{2}-12a=90\] ?

  4. avai70178
    • one year ago
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    then divide -12 by two and square it and add that value to both sides?

  5. avai70178
    • one year ago
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    So that we have \[6a ^{2}-12a+36=126\]

  6. avai70178
    • one year ago
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    That's what I have so far

  7. Astrophysics
    • one year ago
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    Well I think it would be easiest if you first divide everything by 6 so we have \[6a^2-12a-90=0 \implies a^2-2a-15=0\]\[\left( \frac{ -2 }{ 2 } \right)^2 = 1\] we do this to both sides, but you're essentially right it doesn't matter if you bring it to left or right side. So for simplicity in your case lets put it as \[a^2-2a=15\] now we add the 1 to both sides so we get \[a^2-2a+1=16\]

  8. Astrophysics
    • one year ago
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    Now you can easily factor \[a^2-2a+1\]

  9. avai70178
    • one year ago
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    oh okay that way more simple thanks

  10. Astrophysics
    • one year ago
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    Yw :)

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