How do you rationalize the denominator and simplify $\frac{1}{1 + \sqrt{3} - \sqrt{5}}$ $1/{1+sqrt(3)-sqrt(5)}$ ?

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How do you rationalize the denominator and simplify $\frac{1}{1 + \sqrt{3} - \sqrt{5}}$ $1/{1+sqrt(3)-sqrt(5)}$ ?

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Answer 1 · 2018-04-02T10:51:18

Attempt to make the three term denominator two terms..

Explanation:

Multiply the fraction by $\frac{1 + \sqrt{3} + \sqrt{5}}{1 + \sqrt{3} + \sqrt{5}}$ $(1+ sqrt3 + sqrt5)/(1+sqrt3+sqrt5)$ . This is the same as multiplying by 1. This is two eliminate two surds.
The denominator you should get after multiplying the two fractions is
$(1 + \sqrt{3} + \sqrt{5}) \cdot (1 + \sqrt{3} - \sqrt{5})$ $(1+sqrt3+sqrt5) * (1+sqrt3-sqrt5)$ . We can write this as $((1 + \sqrt{3}) - \sqrt{5}) (1 + \sqrt{3}) + \sqrt{5})$ $((1+sqrt3)-sqrt5)(1+sqrt3)+sqrt5)$ .
This can then be made is DOPS and written as : ${(1 + \sqrt{3})}^{2} - {(\sqrt{5})}^{2}$ $(1+sqrt3)^2-(sqrt5)^2$ .
Simplify that to get: $1 + 3 + 2 \sqrt{3} - 5$ $1+3+2sqrt3-5$
I'll leave it here :).

Answer 2 · 2018-04-02T11:17:34

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{7 + 3 \sqrt{3} + \sqrt{5} + 2 \sqrt{15}}{11}$ $1/(1+sqrt(3)-sqrt(5)) =(7+3sqrt(3)+sqrt(5)+2sqrt(15))/11$

Explanation:

This involves two stages of rationalisation to get rid of terms in $\sqrt{3}$ $sqrt(3)$ and $\sqrt{5}$ $sqrt(5)$ . Both steps use the difference of squares identity:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2=(A-B)(A+B)$

So:

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

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{1 + \sqrt{3} + \sqrt{5}}{{(1 + \sqrt{3})}^{2} - {(\sqrt{5})}^{2}}$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = (1+sqrt(3)+sqrt(5))/((1+sqrt(3))^2-(sqrt(5))^2)$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{1 + \sqrt{3} + \sqrt{5}}{1 + 2 \sqrt{3} + 3 - 5}$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = (1+sqrt(3)+sqrt(5))/(1+2sqrt(3)+3-5)$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{1 + \sqrt{3} + \sqrt{5}}{2 \sqrt{3} - 1}$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = (1+sqrt(3)+sqrt(5))/(2sqrt(3)-1)$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{(1 + \sqrt{3} + \sqrt{5}) (2 \sqrt{3} + 1)}{(2 \sqrt{3} - 1) (2 \sqrt{3} + 1)}$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = ((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/((2sqrt(3)-1)(2sqrt(3)+1))$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{(1 + \sqrt{3} + \sqrt{5}) (2 \sqrt{3} + 1)}{{(2 \sqrt{3})}^{2} - 1^{2}}$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = ((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/((2sqrt(3))^2-1^2)$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{(1 + \sqrt{3} + \sqrt{5}) (2 \sqrt{3} + 1)}{12 - 1}$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = ((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/(12-1)$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{1}{11} (1 + \sqrt{3} + \sqrt{5}) (2 \sqrt{3} + 1)$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = 1/11(1+sqrt(3)+sqrt(5))(2sqrt(3)+1)$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{1}{11} (2 \sqrt{3} (1 + \sqrt{3} + \sqrt{5}) + 1 (1 + \sqrt{3} + \sqrt{5}))$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = 1/11(2sqrt(3)(1+sqrt(3)+sqrt(5))+1(1+sqrt(3)+sqrt(5)))$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{1}{11} (2 \sqrt{3} + 6 + 2 \sqrt{15} + 1 + \sqrt{3} + \sqrt{5})$ $color(white)(1/(1+sqrt(3)-sqrt(5))) = 1/11(2sqrt(3)+6+2sqrt(15)+1+sqrt(3)+sqrt(5))$

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{7 + 3 \sqrt{3} + \sqrt{5} + 2 \sqrt{15}}{11}$ $color(white)(1/(1+sqrt(3)-sqrt(5))) =(7+3sqrt(3)+sqrt(5)+2sqrt(15))/11$

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