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Algebra II b











































rationalize the denominator and then simplify 5/2+~3

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Mar 02, 2011
Algebra II b
by: Staff

The question:

rationalize the denominator and then simplify 5/2+~3



The answer:

5/[2+sqrt(3)]


The square root sign in the denominator will disappear if we apply the difference of squares formula.

The difference of squares formula states that:

(a + b)(a − b) = a² − b²



Therefore, if we could multiply the denominator by its conjugate [which is 2-sqrt(3)], the result would contain no square root sign. The denominator would simply = 1:

[2+sqrt(3)]*[2-sqrt(3)] = 2² − [sqrt(3)]²

= 4 - 3

= 1


However, in order to preserve the value of the original fraction, both the numerator and denominator must each be multiplied by the same amount: [2-sqrt(3)].

5/[2+sqrt(3)] must be multiplied by the faction [2-sqrt(3)]/[2-sqrt(3)].

Note that: [2-sqrt(3)]/[2-sqrt(3)] = 1

{5/[2+sqrt(3)]}*1 = {5/[2+sqrt(3)]}*{[2-sqrt(3)]/[2-sqrt(3)]}

= {5*[2-sqrt(3)]}/{[2+sqrt(3)]*[2-sqrt(3)]}

= {5*[2-sqrt(3)]}/{2² − [sqrt(3)]²}

= {5*[2-sqrt(3)]}/1

= 5*[2-sqrt(3)]

= 10 - 5*sqrt(3)

The final answer is: 10 - 5*sqrt(3)


Thanks for writing.


Staff
www.solving-math-problems.com



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