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gfields444

  • 3 years ago

Solve x2 – 4x = 12 by completing the square

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  1. mathslover
    • 3 years ago
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    \[\large{x^2-4x-12=0}\] right?

  2. ProgramGuru
    • 3 years ago
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    you just add 4 to both sides. that will make it \[(x-2)^{2}=16\]

  3. gfields444
    • 3 years ago
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    @mathslover yes

  4. mathslover
    • 3 years ago
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    ok so now .. \[\large{x^2-4x-12=x^2-4x+4-16=0}\] right?

  5. gfields444
    • 3 years ago
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    Yeah

  6. mathslover
    • 3 years ago
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    now can u write \[\large{x^2-4x+4=(x-2)^2}\] ??

  7. mathslover
    • 3 years ago
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    (x)^2 -2(2)(x)+2^2 = a^2-2ab+b^2=(a-b)^2=(x-2)^2

  8. gfields444
    • 3 years ago
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    Stuck

  9. mathslover
    • 3 years ago
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    ok so : (x^2-4x+4) = (x-2)^2 ?

  10. gfields444
    • 3 years ago
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    x^2-4x-12=0 x=-2 x=6? or am I doing it wrong

  11. mathslover
    • 3 years ago
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    I think you are doing it wrong ; : x^2-4x-12 = 0 \[\large{x^2-4x+4-16=0}\] \[\large{(x-2)^2-16=0}\] \[\large{(x-2-4)(x-2+4)=0}\] \[\large{(x-6)(x+2)=0}\] hence either x = 6 or x = -2

  12. mathslover
    • 3 years ago
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    always remember that : \[\large{(a-b)^2=a^2-2ab+b^2}\] \[\large{(a+b)^2=a^2+2ab+b^2}\]

  13. mathslover
    • 3 years ago
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    or : \[\large{(a\pm b)^2=a^2\pm 2ab +b^2}\]

  14. gfields444
    • 3 years ago
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    You're awesome! Thank you so much! The formulas were very helpful :)

  15. mathslover
    • 3 years ago
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    nice to hear .. :) best of luck

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