gfields444
Solve x2 – 4x = 12 by completing the square
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mathslover
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\[\large{x^2-4x-12=0}\] right?
ProgramGuru
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you just add 4 to both sides.
that will make it
\[(x-2)^{2}=16\]
gfields444
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@mathslover yes
mathslover
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ok so now ..
\[\large{x^2-4x-12=x^2-4x+4-16=0}\] right?
gfields444
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Yeah
mathslover
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now can u write \[\large{x^2-4x+4=(x-2)^2}\] ??
mathslover
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(x)^2 -2(2)(x)+2^2 = a^2-2ab+b^2=(a-b)^2=(x-2)^2
gfields444
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Stuck
mathslover
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ok so : (x^2-4x+4) = (x-2)^2 ?
gfields444
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x^2-4x-12=0
x=-2
x=6? or am I doing it wrong
mathslover
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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
mathslover
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always remember that :
\[\large{(a-b)^2=a^2-2ab+b^2}\]
\[\large{(a+b)^2=a^2+2ab+b^2}\]
mathslover
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or : \[\large{(a\pm b)^2=a^2\pm 2ab +b^2}\]
gfields444
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You're awesome! Thank you so much! The formulas were very helpful :)
mathslover
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nice to hear .. :) best of luck