When denominator is rationalised- then the number 3-5-3-5-3 becomes

The correct option is A 9+2√(15) 3√(5)+√(3)/√(5) √(3) #160;=( 3√( 5 ) +√( 3 ) #160;) ( √( 5 ) +√( 3 ) #160;) #160;/( √( 5 ) √( 3 ) #16 ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

When denomina...

Question

When denominator is rationalised, then the number $\frac{3 \sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ becomes

9 + 2 \sqrt{15}

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9 - 2 \sqrt{15}

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9 + 2 \sqrt{9}

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9 - 2 \sqrt{9}

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Solution

The correct option is A $9 + 2 \sqrt{15}$
$\frac{3 \sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$
$= \frac{(3 \sqrt{5} + \sqrt{3}) (\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3}) (\sqrt{5} + \sqrt{3})}$
$= \frac{3 \times 5 + 3 \sqrt{15} + \sqrt{15} + 3}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}}$
$= \frac{15 + 4 \sqrt{15} + 3}{5 - 3}$
$= \frac{18 + 4 \sqrt{15}}{2}$
$= \frac{2 (9 + 2 \sqrt{15})}{2}$
$= (9 + 2 \sqrt{15})$

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Similar questions

[\frac{5}{7} \times \frac{28}{15} \times \frac{9}{8} \times 3] \div \frac{9}{2} =

Q. Rationalise the denominator of

\frac{3 \sqrt[3]{5}}{\sqrt[3]{9}}

Name the property of multiplication illustrated by each of the following statements :

(i) $\frac{- 15}{8} \times \frac{- 12}{7} = \frac{- 12}{7} \times \frac{- 15}{8}$

(ii) $(\frac{- 2}{3} \times \frac{7}{9}) \times \frac{- 9}{5} = \frac{- 2}{3} \times (\frac{7}{9} \times \frac{- 9}{5})$

(iii) $\frac{- 3}{4} \times (\frac{- 3}{4} \times \frac{- 5}{6}) + (\frac{- 3}{4} \times \frac{7}{8})$

(iv) $\frac{- 16}{9} \times 1 = 1 \times \frac{- 16}{9} = \frac{- 16}{9}$

(v) $\frac{- 11}{15} \times \frac{15}{- 11} = \frac{15}{- 11} \times \frac{- 11}{15} = 1$

(vi) $\frac{- 7}{5} \times 0 = 0$

Evaluate : $(i) \frac{5}{9} + \frac{- 7}{6} (i i) 4 + \frac{3}{- 5} (i i i) \frac{1}{- 15} + \frac{5}{- 12} (i v) \frac{5}{9} + \frac{3}{- 4} (v) \frac{- 8}{9} + \frac{- 5}{12} (v i) 0 + \frac{- 2}{7} (v i i) \frac{5}{- 11} + 0 (v i i i) 2 + \frac{- 3}{5} (i x) \frac{4}{- 9} + 1$

A r r a n g e t h e f o l l o w i n g i n a s c e n d i n g o r d e r (1) \frac{2}{9}, \frac{5}{9}, \frac{3}{9}, \frac{5}{9} (2) \frac{3}{5}, \frac{2}{5}, \frac{1}{5}, \frac{4}{5} (3) \frac{4}{7}, \frac{2}{7}, \frac{3}{7}, \frac{6}{7} (4) \frac{10}{15}, \frac{12}{15}, \frac{11}{15}, \frac{15}{15}

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When denominator is rationalised, then the number 3√5+√3√5−√3 becomes

The correct option is A 9+2√153√5+√3√5−√3=(3√5+√3)(√5+√3)(√5−√3)(√5+√3)=3×5+3√15+√15+3(√5)2−(√3)2=15+4√15+35−3=18+4√152=2(9+2√15)2=(9+2√15)

When denominator is rationalised, then the number $\frac{3 \sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ becomes