Question

rationalize 4 + 3 root 5 upon 4 minus 3 root 5

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

To rationalize we must change the dinominator terms. Lets see!

$= \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \\ = \frac{4 + 3\sqrt{5} }{4 - 3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} } \\ = \frac{ {(4 + 3 \sqrt{5} )}^{2} }{ {4}^{2} - {(3 \sqrt{5}) }^{2} } \\ = \frac{ {4}^{2} + {(3 \sqrt{5}) }^{2} + 2(4)(3 \sqrt{5} )}{16 - 9 \times 5} \\ = \frac{16 + 25 + 24 \sqrt{5} }{16 - 25} \\ = \frac{41 + 24 \sqrt{5} }{ - 9} \\ = - \frac{(41 + 24 \sqrt{5} )}{9}$

Hence, we get to answer. Please comment if any doubt is there regaurding question.

$\textbf{ Have great future ahead! }$

Answer 2

HEY MATE HERE IS YOUR ANSWER

$\rule {168}{2}$

$\huge = > \frac{ 4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} }$

$\huge = > \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} }$

$\huge = > \frac{(4 + 3 { \sqrt{5}) }^{2} }{ {4}^{2} - (3 { \sqrt{5} )}^{2} }$

$\huge = > \frac{ {4}^{2} + (3 { \sqrt{2} )}^{2} + 2(4)(3 \sqrt{5}) }{16 - 9 \times 5}$


$\huge = > \frac{16 + 25 + 24 \sqrt{5} }{16 - 25}$

$\huge = > \frac{41 + 24 \sqrt{5} }{ - 9}$

$\huge = > - \frac{(41 + 25 \sqrt{5)} }{9}$

$\rule {135} {2}$

Hope it will help you

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