How do you simplify 7/(5+sqrt3)75+√3?

2 Answers
Veerupaksh · Shwetank Mauria
Mar 31, 2017

=35/22-(7sqrt3)/22=3522−7√322

Explanation:

You must know that the conjugate of an irrational number of the type
a+sqrtba+√b is given by a-sqrtba−√b and that of a-sqrtba−√b is given by a+sqrtba+√b.

7/(5+sqrt3)75+√3

(multiplying and dividing by conjugate of 5+sqrt35+√3)

= 7/(5+sqrt3)*(5-sqrt3)/(5-sqrt3)=75+√3⋅5−√35−√3

= (7(5-sqrt3))/(5^2-(sqrt3)^2)=7(5−√3)52−(√3)2

=(7(5-sqrt3))/22=7(5−√3)22
=35/22-(7sqrt3)/22=3522−7√322

MM
Mar 31, 2017

Rationalise the denominator by multiplying by the conjugate surd.

Explanation:

7/(5+sqrt3)75+√3 = 7/(5 + sqrt3) * (5-sqrt3)/(5-sqrt3)75+√3⋅5−√35−√3

= [7(5-sqrt3)]/[(5^2)-(sqrt3)^2]7(5−√3)(52)−(√3)2

= (35 - 7sqrt3)/(25 - 3)35−7√325−3

= (35 - 7sqrt3)/2235−7√322

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