Pre-Calculus

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I think I need help with this problem-
3+ square root(x-6) = square root of (x+9)
I thought I should square everything to get rid of square root signs
so I had 9 + x-6 = x+ 9
then I would have x+ 3once I subtracted 9-6 from the left side = x+9 on the right side which doesn't make any sense to move anything anymore it doesn't work what did I do wrong? Thanks

  • Pre-Calculus -

    3 + √(x-6) = √(x+9)
    you have to square the whole side, not just each term

    9 + 6√(x-6) + x-6 = x+9
    6√(x-6) = 6
    √(x-6) = 1
    square again
    x-6 = 1
    x = 7

    check: This is MUST, since we squared each side

    LS = 3 + √1 = 4
    RS = √16 = 4

    therefore x = 7

  • Pre-Calculus-Please recheck -

    I'm probably just not understanding yet but Where did the 6 in front of square root of (x-6) come from and where did the 9 disappear to

  • Pre-Calculus -

    left side squared
    =(3+√(x-6) )^2
    = (3+√(x-6))(3+√(x-6))
    = 9 + 3√(x-6) + 3√(x-6) + x-6
    = 9 + 6√(x-6) + x-6

    right side squared = √(x+9) ^2 = x+9

    so
    9 + 6√(x-6) + x-6 = x+9 , collect all terms to right side, leave 6√.... on the left
    6√(x-6) = x-x + 9-9 + 6
    6√(x-6) = 6
    divide both sides by 6
    √(x-6) = 1
    square again
    x-6 = 1
    etc

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