How do you rationalize the denominator #(1+sqrt2)/(3+sqrt5)#?

Question

Julia Adams · Accepted Answer

Multiply both the numerator (top) and denominator (bottom) by the conjugate #3-sqrt(5)#... #(1+sqrt(2))/(3+sqrt(5))# #= (1+sqrt(2))/(3+sqrt(5))*(3-sqrt(5))/(3-sqrt(5))# #= ((1+sqrt(2))(3-sqrt(5)))/((3+sqrt(5))(3-sqrt(5)))# #= ((1+sqrt(2))(3-sqrt(5)))/(3^2-sqrt(5)^2)# #= ((1+sqrt(2))(3-sqrt(5)))/(9-5)# #= ((1+sqrt(2))(3-sqrt(5)))/4#

Nathan Axel · Answer

undefined To rationalize the denominator (1+sqrt2)/(3+sqrt5), we multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of 3+sqrt5 is 3-sqrt5.

By multiplying the numerator and denominator by 3-sqrt5, we get:
[(1+sqrt2)(3-sqrt5)] / [(3+sqrt5)(3-sqrt5)]

Expanding the numerator and denominator, we have:
(3 - sqrt5 + 3sqrt2 - sqrt10) / (9 - 5)

Simplifying further, we get:
(6 + 3sqrt2 - sqrt5 - sqrt10) / 4

Therefore, the rationalized form of (1+sqrt2)/(3+sqrt5) is (6 + 3sqrt2 - sqrt5 - sqrt10) / 4.

Jace Anderson · Answer

undefined To rationalize the denominator of $ \frac{1 + \sqrt{2}}{3 + \sqrt{5}} $, we multiply the numerator and denominator by the conjugate of the denominator.

The conjugate of $ 3 + \sqrt{5} $ is $ 3 - \sqrt{5} $.

So, we multiply the fraction by $ \frac{3 - \sqrt{5}}{3 - \sqrt{5}} $:

$$ \frac{1 + \sqrt{2}}{3 + \sqrt{5}} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}} $$

This gives us:

$$ \frac{(1 + \sqrt{2})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} $$

Expanding the numerator and denominator:

Numerator: $ (1 + \sqrt{2})(3 - \sqrt{5}) = 3 + \sqrt{6} - 3\sqrt{2} - \sqrt{10} $

Denominator: $ (3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4 $

Putting these back into the fraction:

$$ \frac{3 + \sqrt{6} - 3\sqrt{2} - \sqrt{10}}{4} $$

This is the rationalized form of the given fraction.