absolute value

|x^2-10|=4
x^2-10=4 or x^2-10=-4
X^2-14=0 x^2-6=0
(x-7)(x+2)=0 (x-2)(x+3)=0
x=7 x=-2 x=2 x=-3
solution set {-3,-2,2,7} (?????)

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asked by Kally
  1. Not quite. You cannot factor x^2-14=0 and get (x-7)(x+2)=0 nor can you factor x^2-6=0 and get (x-2)(x+3). Try to expand them and you'll see they are not the same.

    (x-7)(x+2) = x^2-5x-14
    (x-2)(x+3) = x^2+x-6

    Your work was find until you tried to factor them. There is a much simpler(?) solution. Try to do it and I'll be happy to check your answer if needed.

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  2. Is this right?

    x^2-14=0 x^2-6=0
    x^2=14 x^2=6
    sqrtx^2=sqrt14 sqrtx^2=sqrt6
    x=+-sqrt14 x=+-sqrt6
    solution set is:
    {+-sqrt6, +-sqrt14}

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    posted by Kally
  3. Yes.

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