What is -5 - -3-5 - -3 - -5 - -3-5 - -3 equal to-

The correct option is B 8 We need to factorize √(5) + #160;√(3)√(5) #160;√(3) + √(5) #160;√(3)√(5) + #160;√(3) = (√(5) + #160;√(3) ...

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Byju's Answer

Standard X

Mathematics

Arithmetic Progression

What is √5 ...

Question

What is $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ equal to?

16

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8

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4

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\sqrt{15}

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Solution

The correct option is B $8$
We need to factorize $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$
$= \frac{(\sqrt{5} + \sqrt{3})^{2} + (\sqrt{5} - \sqrt{3})^{2}}{(\sqrt{5})^{2} - (\sqrt{3})^{2}}$
$= \frac{5 + 3 + 2 \sqrt{15} + 5 + 3 - 2 \sqrt{15}}{2}$
$= \frac{16}{2} = 8$

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Similar questions

Q. The value of

[\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \frac{1}{\sqrt{5} + \sqrt{4}} + . . . . + \frac{1}{\sqrt{16} + \sqrt{15}}]

is equal to

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

$Name the multiplication property of rational numbers shown below: (i) \frac{3}{5} \times \frac{- 8}{9} = \frac{- 8}{9} \times \frac{3}{5} (i i) \frac{- 3}{4} \times (\frac{5}{7} \times \frac{- 8}{15}) = (\frac{- 3}{5} \times \frac{5}{7}) \times \frac{- 8}{15} (i i i) \frac{4}{5} \times (\frac{3}{- 8} + \frac{- 4}{7}) = \frac{4}{5} \times \frac{3}{- 8} + \frac{4}{5} \times \frac{- 4}{7} (i v) \frac{- 7}{5} \times \frac{5}{- 7} = 1 (v) \frac{8}{- 9} \times 1 = 1 \times \frac{8}{- 9} = \frac{8}{- 9} (v i) \frac{- 3}{4} \times 0 = 0$