Multiplication and Division of Radicals

Key Questions

  • #sqrt(a/b)= [sqrta]/[sqrtb]#

    #sqrta xx sqrtb=sqrt(axxb)#

    or #sqrtasqrtb=sqrt(axxb)#

  • When you have a root (square root for example) in the denominator of a fraction you can "remove" it multiplying and dividing the fraction for the same quantity. The idea is to avoid an irrational number in the denominator.
    Consider:
    #3/sqrt2#
    you can remove the square root multiplying and dividing by #sqrt2#;
    #3/sqrt2*sqrt2/sqrt2#
    This operation does not change the value of your fraction because #sqrt2/sqrt2=1# anyway and your fraction does not change by multiplying #1# to it.

    Now you can multiply in the numerator and denominator:
    #3/sqrt2*sqrt2/sqrt2=(3*sqrt2)/((sqrt2)*(sqrt2))# giving:
    #(3*sqrt2)/2# you have removed the square root from the denominator! (ok it went to the nominator but this is ok).

    Now a problem for you; what happens when the root is not alone???!!!
    If you have:
    #3/(1+sqrt2)#???
    You can use the same technique but...what do you use to multiply and divide?
    HINT: look at what happens if you do this:
    #(1+sqrt2)*(1-sqrt2)#

  • Multiplication

    #sqrt{a}cdot sqrt{b}=sqrt{a cdot b}#

    Division

    #{sqrt{a}}/{sqrt{b}}=sqrt{a/b}#


    I hope that this was helpful.

Questions