Posted by sue on Monday, October 25, 2010 at 9:00pm.
when you have a term squareroot (for example sqrt(x))in the denominator, multiply it with itself such that the multiplier is equal to one ( sqrt(x)/sqrt(x) ) to remove the squareroot,, in this problem:
1+(-2/(sqroot 5))
1+(-2/(sqroot 5))*(sqrt(5)/sqrt(5))
1+{[-2*sqrt(5)]/5}
therefore:
{[5-2*sqrt(5)]/5}
so there,, :)
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