sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =

Question

\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} =

A.  4 + \sqrt{15} 
B. 4
C.  \sqrt{15} 
D.  4 - \sqrt{15} 
E.

PareshGmat · Accepted Answer

Multiply numerator & denominator by  (\sqrt{5} + \sqrt{3})

= \frac{(\sqrt{5} + \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}

= \frac{5 + 2\sqrt{15} + 3}{5 - 3}

= \frac{8 + 2\sqrt{15}}{2}

= 4 + \sqrt{15}

Formulae used:

(a+b)^2 = a^2 + 2ab + b^2

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

duttsit · Answer

(sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3))

multiple both numerator and denominator by :
(sqrt(5)+sqrt(3))

gives:

{  ^ 2} / 2

or 

4 + sqrt (15)

chets · Answer

the reason i posted this question was to ask a second question.

if you chose to multiply the numerator and denominator by sqrt(5) + sqrt(3) you get 4+sqrt(5)*sqrt(3)

though, if you chose to multiply the numerator and denominator by sqrt(5) - sqrt(3) you get 1 / (4-sqrt(5)*sqrt(3))

though seemingly different, 

4+sqrt(5)*sqrt(3) = 1/(4-sqrt(5)*sqrt(3))

I am not sure whether there is a broad generalization i am missing. If anyone can shed light on this, that would be great!

krisrini · Answer

Take conjugate on both sides. which gives.

(sqrt(5) + sqrt(3)) (sqrt(5) + sqrt(3)) / (sqrt(5) - sqrt(3)) (sqrt(5) + sqrt(3))

This gives (sqrt(5) +sqrt(3))^2 / (Sqrt(5)^2) - (sqrt(3))^2

=> 5 + 3 + 2 sqrt(15) / 2

=> 2(4 + sqrt(15))/2

=> 4 + sqrt(15)

Thanks

macca · Answer

Chet, You r not missing anything, You get: 1/(4-sqrt(5)*sqrt(3)) = (4+sqrt(5)*sqrt(3)) / = (4+sqrt(5)*sqrt(3)) / (16 - 15) = (4+sqrt(5)*sqrt(3)) Hope that helps

kundankshrivastava · Answer

This type of questions can be solved by using formula

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

Denominator is in the form of a-b , hence we will multiply numerator and denominator by the form a +b .i.e. sqrt(5) + sqrt(3)

sqrt(5) + sqrt(3)/sqrt(5) - sqrt(3)

* / *

5+3 + 2*sqrt(5)*sqrt(3) / 5-3

8+2*sqrt(5)*sqrt(3)/2

4+sqrt(15)

Bunuel · Answer

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JapinderKaur · Answer

Chets asked if 4+sqrt(5)*sqrt(3) = 1/(4-sqrt(5)*sqrt(3)) can be extrapolated into a broad generalization. No, it can't be!

Here's why.

Consider an expression  1/(a-\sqrt{bc})   
Multiplying Numerator and Denominator with  (a+ \sqrt{bc}) , we get
 (a+ \sqrt{bc})/(a^2-bc)

So, the reason why 1/(4-sqrt(5)*sqrt(3) got simplified into 4+sqrt(5)*sqrt(3) was because the Denominator of the above expression is equal to 1 in this particular case (a=4, b=5, c=3. So, a^2-bc = 4^2-5*3 = 1 )

Emily154 · Answer

the answer is  4 + \sqrt{15}

BrushMyQuant · Answer

↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧

https://www.youtube.com/watch?v=2QSbTwQ_G2o

We need to simplify  \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}

This problem is based on Rationalizing Roots. We will simplify the denominator by multiply the numerator and the denominator with its conjugate  {\sqrt{5}+\sqrt{3}}

=>  \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}  *  \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}

Now the denominator becomes (a-b) * (a+b) and will be equal to  a^2 - b^2

=>  \frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5})^2-(\sqrt{3})^2}  =  \frac{5 + 2*\sqrt{5}*\sqrt{3} + 3}{5 - 3}  =  \frac{8 + 2\sqrt{15}}{2}  = 4 +  \sqrt{15}

So,  Answer will be 4 +  \sqrt{15}  
Hope it helps!

Watch the following video to learn How to Rationalize Roots

https://www.youtube.com/watch?v=LhR_gXt8CLw

bumpbot · Answer

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sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =

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sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =

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