Math, asked by wwwsrishtiyadavcom, 2 months ago

find the values of a and b​

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Answered by Anonymous
23

Question :-

\sf \dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3} = a + b\sqrt{15}

Answer :-

Solving LHS :-

\implies\sf \dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3} \times \dfrac{\sqrt5 + \sqrt3}{\sqrt5 +\sqrt3}

\implies\sf \dfrac{(\sqrt5 + \sqrt3)(\sqrt5 + \sqrt3)}{(\sqrt5 - \sqrt3)(\sqrt5 + \sqrt3)}

\implies\sf \dfrac{(\sqrt5 + \sqrt3)^2}{(\sqrt5 - \sqrt3)(\sqrt5 + \sqrt3)}

  • \sf (a+b)^2 = a^2 + b^2 + 2ab
  • \sf (a-b)(a+b) = a^2 - b^2

\implies\sf \dfrac{(\sqrt5)^2 + (\sqrt3)^2 + 2 \times \sqrt5 \times sqrt3}{(\sqrt5)^2 - (\sqrt3)^2}

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

\implies\sf \dfrac{8+2\sqrt{15}}{2}

\implies\sf 4 + \sqrt{15}

Comparing with RHS :-

\implies\sf a + b\sqrt{15} = 4 + \sqrt{15}

  • a = 4
  • b = 1

Value of a = 4 & b = 1

Answered by emma3006
1

Answer:

a = 4

b = 1

Step-by-step explanation:

Given:

\sf{\;\;\;\;\;\;\dfrac{\sqrt{5}+ \sqrt{3}}{\sqrt{5}-\sqrt{3}} = a+b\sqrt{15} }

To find:

Find the values of a and b

Identities:

  1. (a+b)² = a²+b²+2ab
  2. a²-b² = (a+b) (a-b)

Solution:

\sf{\;\;\;\;\;\;\dfrac{\sqrt{5}+ \sqrt{3}}{\sqrt{5}-\sqrt{3}} = a+b\sqrt{15} }

Rationalizing the denominator,

\implies \sf{\dfrac{\sqrt{5}+ \sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \dfrac{\sqrt{5}+ \sqrt{3}}{\sqrt{5}+\sqrt{3}}= a+b\sqrt{15} }

\implies \sf{\dfrac{(\sqrt{5}+ \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2} = a+b\sqrt{15} }

Solving the numerator using identity (1) and denominator using the identity (2)

\implies \sf{\dfrac{(\sqrt{5})^{2} + (\sqrt{3})^{2} +2 \times \sqrt{5} \times \sqrt{3}}{5-3} = a+b\sqrt{15} }

\implies \sf{\dfrac{5+3+2\sqrt{15}}{2} = a+b\sqrt{15} }

\implies \sf{\dfrac{8+2\sqrt{15}}{2} = a+b\sqrt{15} }

\implies \sf{4+\sqrt{15} = a+b\sqrt{15} }

Comparing LHS and RHS, we get

a = 4

b = 1

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