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Algebra

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Square root of (q^2+7q+6) minus q minus 3=0. Minus 3 and minus q are not under he radical sign.

  • Algebra - ,

    I assume your problem is this:

    √(q^2 + 7

  • Algebra - ,

    Oops! Let's try this again.

    I assume your problem is this:

    √(q^2 + 7q + 6) - q - 3 = 0

    Add q and 3 to both sides of the equation to get the radical by itself (whatever operation you do to one side of an equation you must do to the other side as well):

    √(q^2 + 7q + 6) = q + 3

    Now square both sides to get rid of the radical:

    q^2 + 7q + 6 = (q + 3)^2

    q^2 + 7q + 6 = q^2 + 6q + 9

    q = 3

    Test the answer with the original equation to see if it checks out. It always helps to check your work!

    I hope this helps and is what you were asking.

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