What is the simplified form of #(sqrt(2)-sqrt(10))/(sqrt(2)+sqrt(10))# ?

2 Answers
George C.
May 24, 2017

#-3/2+1/2sqrt(5)#

None of the given options is correct.

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Hence we can rationalise the denominator of the given expression by multiplying both numerator and denominator by #sqrt(2)-sqrt(10)# as follows:

#(sqrt(2)-sqrt(10))/(sqrt(2)+sqrt(10)) = (sqrt(2)-sqrt(10))^2/((sqrt(2)-sqrt(10))(sqrt(2)+sqrt(10)))#

#color(white)((sqrt(2)-sqrt(10))/(sqrt(2)+sqrt(10))) = ((sqrt(2))^2-2sqrt(2)sqrt(10)+(sqrt(10))^2)/((sqrt(2))^2-(sqrt(10))^2)#

#color(white)((sqrt(2)-sqrt(10))/(sqrt(2)+sqrt(10))) = (2-2sqrt(20)+10)/(2-10)#

#color(white)((sqrt(2)-sqrt(10))/(sqrt(2)+sqrt(10))) = (12-2sqrt(2^2*5))/(-8)#

#color(white)((sqrt(2)-sqrt(10))/(sqrt(2)+sqrt(10))) = (12-4sqrt(5))/(-8)#

#color(white)((sqrt(2)-sqrt(10))/(sqrt(2)+sqrt(10))) = -3/2+1/2sqrt(5)#

Tazwar Sikder
May 24, 2017

#- frac(3)(2) + frac(1)(2) sqrt(5)#

Explanation:

We have: #frac(sqrt(2) - sqrt(10))(sqrt(2) + sqrt(10))#

Let's rationalise the denominator:

#= frac(sqrt(2) - sqrt(10))(sqrt(2) + sqrt(10)) cdot frac(sqrt(2) - sqrt(10))(sqrt(2) - sqrt(10))#

#= frac((sqrt(2) + sqrt(10))^(2))((sqrt(2))^(2) - (sqrt(10))^(2))#

#= frac(2 - 2 sqrt(20) + 10)(2 - 10)#

#= frac(12 - 2 sqrt(2^(2) cdot 5))(- 8)#

#= frac(12 - 2 cdot 2 sqrt(5))(- 8)#

#= frac(12 - 4 sqrt(5))(- 8)#

#= - frac(6 - 2 sqrt(5))(4)#

#= - frac(3 - sqrt(5))(2)#

#= - frac(3)(2) + frac(1)(2) sqrt(5)#

It seems that none of the options match this.

Perhaps you made an error while typing the possible answers.

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