Expert: Scotto - 7/15/2009

Question
Can u help to solve this problems

a)1/root10+root14+root15+root21 with rational denominator
b)8th root of 3+8th root of 2 with rationalising factor

Answer
a) Whenever I see numbers that small, the factors automatically click in my head.  I know that 10=2*5, 14=2*7, 15=3*5, and 21=3*7.

I will use rt() for root for this problem.

Once this is seen, notice that the numbers 2, 3, 5, and 7 all occur in two numbers.  This means that (rt(2)+rt(3))(rt(5)+rt(7)) is the same as rt(10)+rt(14)+rt(15)+rt(21).

If we multiply the numerator and denominator by
(rt(3)-rt(2))(rt(7)-rt(5)), we get

(rt(3)-rt(2))(rt(7)-rt(5))/
(rt(2)+rt(3))(rt(5)+rt(7))(rt(3)-rt(2))(rt(7)-rt(5))

where the second line is all in the denominator.

Now it should be noted that the denominator can be rearranged into
(rt(2)+rt(3))(rt(3)-rt(2))(rt(5)+rt(7))(rt(7)-rt(5)).

Effectively, recognize that the function is (b+a)(b-a)(d+c)(c-d).
The result is then (b^2-a^2)(d^2-c^2).
Next note in the result, all of the numbers are squared.

This gives us the real answer is
(rt(3)-rt(2))(rt(7)-rt(5))/(3-2)(7-5) =
(rt(3)-rt(2))(rt(7)-rt(5))/(1*2) =
(rt(3)-rt(2))(rt(7)-rt(5))/2.


b)Let the nth root of a be nrt(a) for this problem.
This makes the problem into 8rt(3)+8rt(2).
That looks more concise.

Now if we multiply by 8rt(3)-8rt(2), we get 8rt(3)^2 - 8rt(2)^2 =
4rt(3) - 4rt(2).  If we multiply that by 4rt(3) + 4rt(2), we get
rt(3) - rt(2).  If we multiply that by rt(3)+rt(2), we get 3-2.

In essence we took 8rt(3)+8rt(2) and multiplied by
(8rt(3)-8rt(2)(4rt(3)+4rt())(rt(3)+rt(2)) and got 1.

Is this what is being looked for?

If the value 8rt(3)+8rt(2) was looked for in one term, it would be
1/(8rt(3)-8rt(2)(4rt(3)+4rt())(rt(3)+rt(2)).