Question

If a = √10+ √ 5 ÷ √ 10 -√5 and b= √10-√ 5÷ √10 +√5 ; then show that √a-√ b- 2√ ab= 0


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Answer 1

Answer:

Step-by-step explanation:

$a = \frac{\sqrt{10} + \sqrt{5} }{\sqrt{10} -\sqrt{5} } \\b = \frac{\sqrt{10} - \sqrt{5} }{\sqrt{10} +\sqrt{5} } \\To Show that\\\sqrt{a} - \sqrt{b} - 2\sqrt{ab} = 0\\\sqrt{a} - \sqrt{b} - 2\sqrt{ab} = 0\\=> \sqrt{a} - \sqrt{b} = 2\sqrt{ab} \\Squaring both sides\\=> a + b - 2\sqrt{ab} = 4ab$

$a = \frac{\sqrt{10} + \sqrt{5} }{\sqrt{10} - \sqrt{5} } \\= \frac{\sqrt{10} + \sqrt{5} }{\sqrt{10} - \sqrt{5} } \times \frac{\sqrt{10} + \sqrt{5} }{\sqrt{10} + \sqrt{5} }\\=\frac{10 + 5 + 2\sqrt{50} }{10-5} \\=\frac{15 + 2\times5\sqrt{2} }{5} \\= 3 + 2\sqrt{2}$

$b = \frac{\sqrt{10} - \sqrt{5} }{\sqrt{10} +\sqrt{5}}\\b = \frac{\sqrt{10} - \sqrt{5} }{\sqrt{10} +\sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5} }{\sqrt{10} -\sqrt{5}}\\b = \frac{10 + 5 - 2\sqrt{50} }{10 -5} \\b = 3 - 2\sqrt{2}$

$ab = (3 + 2\sqrt{2})(3 - 2\sqrt{2})\\ab = 9 -8\\ab = 1$

putting values of a & b in

a + b - 2√ab = 4ab

=> 3 +2√2 + 3 -2√2 - 2√1 = 4×1

=> 6 - 2 = 4

=> 4 - 4 = 0

as √0 = 0

hence √a - √b - 2√ab = 0

QED

Answer 2

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