How do you rationalize (2sqrt5-8)/ (2sqrt5+3)25825+3?

2 Answers
Mar 4, 2018

2(2-sqrt5)2(25)

Explanation:

(2 sqrt5-8)/(2sqrt5+3)25825+3. Multiplying by (2sqrt5-3)(253) on

both numerator and denominator we get,

=((2 sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))=(258)(253)(25+3)(253)

=(20-2sqrt5(8+3)+24)/((2sqrt5)^2-3^2)=2025(8+3)+24(25)232

=(44-22sqrt5)/(20-9)=(22(2-sqrt5))/11=44225209=22(25)11

=2(2-sqrt5)=2(25) [Ans]

Mar 4, 2018

(2sqrt5-8)/(2sqrt5+3)=4-2sqrt525825+3=425

Explanation:

To rationalize the denominator, we multiply by the conjugate and use the difference of squares rule. In this case, the conjugate is 2sqrt5-3253, so we multiply by it on both top and bottom:

(2sqrt5-8)/(2sqrt5+3)=((2sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))25825+3=(258)(253)(25+3)(253)

The difference of squares rule says:
(a+b)(a-b)=a^2-b^2(a+b)(ab)=a2b2

Applying this to the denominator, we get:
((2sqrt5-8)(2sqrt5-3))/(4*5-3)(258)(253)453

Then we multiply out the top:
(20-6sqrt5-16sqrt5+24)/11=(44-22sqrt5)/11=4-2sqrt52065165+2411=4422511=425