How do you rationalize (2sqrt5-8)/ (2sqrt5+3)2√5−82√5+3?

2 Answers
Binayaka C.
Mar 4, 2018

2(2-sqrt5)2(2−√5)

Explanation:

(2 sqrt5-8)/(2sqrt5+3)2√5−82√5+3. Multiplying by (2sqrt5-3)(2√5−3) on

both numerator and denominator we get,

=((2 sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))=(2√5−8)(2√5−3)(2√5+3)(2√5−3)

=(20-2sqrt5(8+3)+24)/((2sqrt5)^2-3^2)=20−2√5(8+3)+24(2√5)2−32

=(44-22sqrt5)/(20-9)=(22(2-sqrt5))/11=44−22√520−9=22(2−√5)11

=2(2-sqrt5)=2(2−√5) [Ans]

Alvin L.
Mar 4, 2018

(2sqrt5-8)/(2sqrt5+3)=4-2sqrt52√5−82√5+3=4−2√5

Explanation:

To rationalize the denominator, we multiply by the conjugate and use the difference of squares rule. In this case, the conjugate is 2sqrt5-32√5−3, so we multiply by it on both top and bottom:

(2sqrt5-8)/(2sqrt5+3)=((2sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))2√5−82√5+3=(2√5−8)(2√5−3)(2√5+3)(2√5−3)

The difference of squares rule says:
(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a2−b2

Applying this to the denominator, we get:
((2sqrt5-8)(2sqrt5-3))/(4*5-3)(2√5−8)(2√5−3)4⋅5−3

Then we multiply out the top:
(20-6sqrt5-16sqrt5+24)/11=(44-22sqrt5)/11=4-2sqrt520−6√5−16√5+2411=44−22√511=4−2√5

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