Question

Rationalize the Denominator of each of the following.
(a) $\bold {\frac{\sqrt{3} - \sqrt{5}}{\sqrt{3} + \sqrt {5}}}$
_____
CLUE : Rationalizing Factor = $\bold {\sqrt {3} - \sqrt {5}}$

Answer 1

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Rationalize the denominator of each of the following

(a) $\bf\dfrac{\sqrt3 - \sqrt5}{\sqrt3 + \sqrt5}$

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⏣ Multiply both numerator as well as denominator of the fraction with the conjugate pair.

⏣ A pair of conjugates is a pair of binomials that are exactly the same except that the signs between the terms are opposite. To create a conjugate of a binomial, just rewrite it and change the sign of the second term.

✦ Example : The conjugate pair of (a + b) is (a - b)

⏣ Need to use some Algebraic Identities :-

✦ (a + b)² = a² + b² + 2ab

✦ (a - b)² = a² + b² - 2ab

✦ (a + b) (a - b) = a² - b²

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$\sf\dfrac{\sqrt3 - \sqrt5}{\sqrt3 + \sqrt5}$

$\sf = \dfrac{\sqrt3 - \sqrt5}{\sqrt3 + \sqrt5} * \dfrac{\sqrt3 - \sqrt5}{\sqrt3 - \sqrt5} \; \;[Multipling\; both \;numerator \;and \;denominator\; with\; conjugate \;pair\; of \;(\sqrt3 + \sqrt5) \; ie.\; (\sqrt3-\sqrt5) ]$

$\sf = \dfrac{(\sqrt3 - \sqrt5) \; (\sqrt3-\sqrt5) }{(\sqrt3 + \sqrt5) \; (\sqrt3-\sqrt5) }$

$\sf = \dfrac{(\sqrt3)^2 +(\sqrt5)^2 -(2*\sqrt3*\sqrt5)}{(\sqrt3)^2 - (\sqrt5)^2}\;\;[Using \; (a-b)^2 = a^2+b^2-2ab\; and \;(a + b)(a - b)= a^2-b^2 ]$

$\sf = \dfrac{3+5-2\sqrt{15}}{3-5}$

$\sf = \dfrac{8 - 2\sqrt{15}}{-2}$

$\sf = \dfrac{-(8 - 2\sqrt{15})}{2}$

$\sf = \dfrac{-8+2\sqrt{15}}{2}$

$\sf = \dfrac{2\sqrt{15}-8}{2}$

∴ The rationalized from of $\bf\dfrac{\sqrt3 - \sqrt5}{\sqrt3 + \sqrt5}$   is $\bf\dfrac{2\sqrt{15}-8}{2}$ after rationalizing the denominator.

Answer 2

Step-by-step explanation:

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rationalize:

$\frac{ \sqrt{3} - \sqrt{5} }{ \sqrt{3} + \sqrt{5} }$

$\frac{ \sqrt{3} - \sqrt{5} }{ \sqrt{3} + \sqrt{5} } \times \frac{ \sqrt{3} - \sqrt{5} }{ \sqrt{3} - \sqrt{5} }$

$\frac{( \sqrt{3} - \sqrt{5}) {}^{2} }{3 - 5}$

$\frac{3 + 5 - 2 \sqrt{15} }{ - 2}$

$\frac{8 - 2 \sqrt{15} }{ - 2}$