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     (2x-3)^2 =9 .

    Simple enough, I managed to deduce that  x=3 .

    What I don't understand, however, is how  x can also equal  0 in this instance, because if  2x-3=3 , and  x=0 , that's  2(0) -3 , which seems to equal  -3 , rather than  3

    Could someone explain how  x could be  0 in this instance?
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    (Original post by Illidan2)
     (2x-3)^2 =9 .

    Simple enough, I managed to deduce that  x=3 .

    What I don't understand, however, is how  x can also equal  0 in this instance, because if  2x-3=3 , and  x=0 , that's  2(0) -3 , which seems to equal  -3 , rather than  3

    Could someone explain how  x could be  0 in this instance?
    when you square root both sides you get 2x-3 = +/- 3
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    Ahh yes. Of course. I always seem to overlook that. I must remember that the square root may also be the negative variant. Thank you.
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    (Original post by Illidan2)
    Ahh yes. Of course. I always seem to overlook that. I must remember that the square root may also be the negative variant. Thank you.
    One thing,  \sqrt{x^2} = |x| \neq -x unless x<0.
    |x| just means the positive value.
    So  \sqrt{9} = 3, however -3 is a solution to the equation.

    This is important, because in later questions, you can't just square root a value and say it is equal to 2 different values.
 
 
 
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