Question

if a=3-root 5/3+root5 ,b=3+root5/3-root 5find a^2 -b^2

Answer 1

$<b>Hey mate!!$

$\bold{a = \frac{3 - \sqrt{5} }{3 + \sqrt{5} } } \\ \\ \bold{a = \frac{3 - \sqrt{5} }{3 + \sqrt{5}} \times \frac{3 - \sqrt{5} }{3 - \sqrt{5} } } \\ \\ \bold{a = \frac{(3 - \sqrt{5}) {}^{2} }{(3) {}^{2} - ( \sqrt{5}) {}^{2} } } \\ \\ \bold{a = \frac{9 - 6 \sqrt{5} + 5}{9 - 5} } \\ \\ \bold{a = \frac{14 - 6 \sqrt{5}}{4} } \\ \\ \bold{a = \frac{\cancel{2}(7 - 3 \sqrt{5} )}{\cancel{4} \: \: 2} } \\ \\ \bold {a = \frac{7 - 3 \sqrt{5} }{2} }$

And,

$\bold{b= \frac{3 + \sqrt{5} }{3 - \sqrt{5} } } \\ \\ \bold{b= \frac{3 + \sqrt{5} }{3 - \sqrt{5}} \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } } \\ \\ \bold{b = \frac{(3 + \sqrt{5}) {}^{2} }{(3) {}^{2} - ( \sqrt{5}) {}^{2} } } \\ \\ \bold{b = \frac{9 + 6 \sqrt{5} + 5}{9 - 5} } \\ \\ \bold{b = \frac{14 + 6 \sqrt{5}}{4} } \\ \\ \bold{b= \frac{\cancel{2}(7 - 3 \sqrt{5} )}{\cancel{4} \: \: 2} } \\ \\ \bold {b= \frac{7 + 3 \sqrt{5} }{2} }$

Now,

$\bold{a {}^{2} - b {}^{2} } \\ \\ \bold {= >( \frac{7 - 3 \sqrt{5} }{2} ) {}^{2} - (\frac{7 + 3 \sqrt{5} }{2} ) {}^{2} } \\ \\ \bold{ = > \frac{49 - 45}{4} - \frac{49 + 44}{4} } \\ \\ \bold{ = > \frac{ \cancel{4}}{ \cancel{4}} - \frac{ \cancel{4}}{ \cancel{4}} } \\ \\ \bold{ 1 - 1 = 0}$

Thanks for the question!

Answer 2

Answer:

see the above pic

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