Question

Find a and b, if ( 5+√3 /5-√3) -(5 -√3/ 5+√3)=a+b√3
5-3
5+63​

Answer 1

Answer:

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Answer 2

Question:

$Find \:\: a \:\: and \:\: b \: \:if \:; \\ \frac{5 + \sqrt{3} }{5 - \sqrt{3} } - \frac{5 - \sqrt{3} }{5 + \sqrt{3} } = a + b \sqrt{3} .$

Answer:

$a = 0 \\ b = \frac{10}{11}$

Note:

• (A+B)² = A² + B² + 2•A•B

• (A-B)² = A² + B² - 2•A•B

• (A+B)•(A-B) = A² - B²

• (A+B)² + (A-B)² = 2(A² + B²)

• (A+B)² - (A-B)² = 4•A•B

Solution:

$= > \frac{5 + \sqrt{3} }{5 - \sqrt{3} } - \frac{5 - \sqrt{3} }{5 + \sqrt{3} } = a + b \sqrt{3} \\ = > \frac{ {(5 + \sqrt{3} })^{2} - {(5 - \sqrt{3}) }^{2} }{(5 - \sqrt{3})(5 + \sqrt{3}) } = a + b \sqrt{3} \\ = > \frac{4 \times 5 \times \sqrt{3} }{ {5}^{2} - { (\sqrt{3}) }^{2} } = a + b \sqrt{3} \\ = > \frac{20 \sqrt{3} }{25 - 3} = a + b \sqrt{3} \\ = > \frac{20 \sqrt{3} }{22} = a + b \sqrt{3} \\ = > \frac{10 \sqrt{3} }{11} = a + b \sqrt{3} \\ = > 0 + \frac{10 \sqrt{3} }{11} = a + b \sqrt{3} \\ \\ Now, \\Comparing \: both \: the \: sides ,\: we \: have ;\: \\ a = 0 \: and \: b = \frac{10}{11}$