To simplify the expression 3 + 1 3 − 1 \frac{\sqrt{3}+1}{\sqrt{3}-1} 3 − 1 3 + 1 , we need to rationalize the denominator. This can be done by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of 3 − 1 \sqrt{3}-1 3 − 1 is 3 + 1 \sqrt{3}+1 3 + 1 . Thus, we have:
3 + 1 3 − 1 = 3 + 1 3 − 1 ⋅ 3 + 1 3 + 1 \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1} 3 − 1 3 + 1 = 3 − 1 3 + 1 ⋅ 3 + 1 3 + 1
Multiplying the numerators gives:
( 3 + 1 ) ( 3 + 1 ) = ( 3 ) 2 + 2 3 + 1 = 3 + 2 3 + 1 = 4 + 2 3 (\sqrt{3}+1)(\sqrt{3}+1) = (\sqrt{3})^2 + 2\sqrt{3} + 1 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} ( 3 + 1 ) ( 3 + 1 ) = ( 3 ) 2 + 2 3 + 1 = 3 + 2 3 + 1 = 4 + 2 3
Multiplying the denominators gives:
( 3 − 1 ) ( 3 + 1 ) = ( 3 ) 2 − 1 2 = 3 − 1 = 2 (\sqrt{3}-1)(\sqrt{3}+1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2 ( 3 − 1 ) ( 3 + 1 ) = ( 3 ) 2 − 1 2 = 3 − 1 = 2
So we have:
3 + 1 3 − 1 = 4 + 2 3 2 \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{4 + 2\sqrt{3}}{2} 3 − 1 3 + 1 = 2 4 + 2 3 Now, divide both terms in the numerator by 2:
4 + 2 3 2 = 4 2 + 2 3 2 = 2 + 3 \frac{4 + 2\sqrt{3}}{2} = \frac{4}{2} + \frac{2\sqrt{3}}{2} = 2 + \sqrt{3} 2 4 + 2 3 = 2 4 + 2 2 3 = 2 + 3 