Question

If P + √3Q + √5R + √15S = 1/1+√3+√5, then find the value of P is
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Answer 1

Step-by-step explanation:

$\mathsf{Consider :\;\dfrac{1}{1 + \sqrt{3} + \sqrt{5}}}$

$\mathsf{Multiplying\;numerator\;and\;denominator\;with\;1 + \sqrt{3} - \sqrt{5} :}$

$\mathsf{\implies \dfrac{1 + \sqrt{3} - \sqrt{5}}{(1 + \sqrt{3} + \sqrt{5})(1 + \sqrt{3} - \sqrt{5})}}$

★  We know that : (a + b)(a - b) = a² - b²

$\mathsf{\implies \dfrac{1 + \sqrt{3} - \sqrt{5}}{(1 + \sqrt{3})^2 - (\sqrt{5})^2}}$

★  We know that : (a + b)² = a² + b² + 2ab

$\mathsf{\implies \dfrac{1 + \sqrt{3} - \sqrt{5}}{1 + (\sqrt{3})^2 + 2\sqrt{3} - 5}}$

$\mathsf{\implies \dfrac{1 + \sqrt{3} - \sqrt{5}}{1 + 3 + 2\sqrt{3} - 5}}$

$\mathsf{\implies \dfrac{1 + \sqrt{3} - \sqrt{5}}{2\sqrt{3} - 1}}$

$\mathsf{Multiplying\;numerator\;and\;denominator\;with\;2\sqrt{3} + 1 :}$

$\mathsf{\implies \dfrac{(1 + \sqrt{3} - \sqrt{5})(2\sqrt{3} + 1)}{(2\sqrt{3} - 1)(2\sqrt{3} + 1)}}$

$\mathsf{\implies \dfrac{(2\sqrt{3} + 1 + (\sqrt{3})(2\sqrt{3}) + \sqrt{3} - (\sqrt{5})(2\sqrt{3}) - \sqrt{5})}{(2\sqrt{3})^2 - 1}}$

$\mathsf{\implies \dfrac{(3\sqrt{3} + 1 + 2(\sqrt{3})^2 - 2(\sqrt{5\times3}) - \sqrt{5})}{4(\sqrt{3})^2 - 1}}$

$\mathsf{\implies \dfrac{3\sqrt{3} + 1 + 2(3) - 2\sqrt{15} - \sqrt{5}}{(4)(3) - 1}}$

$\mathsf{\implies \dfrac{3\sqrt{3} + 1 + 6 - 2\sqrt{15} - \sqrt{5}}{12 - 1}}$

$\mathsf{\implies \dfrac{7 + 3\sqrt{3} - \sqrt{5} - 2\sqrt{15}}{11}}$

$\mathsf{\implies \dfrac{7}{11} + \dfrac{3\sqrt{3}}{11} - \dfrac{\sqrt{5}}{11} - \dfrac{2\sqrt{15}}{11}}$

$\mathsf{\implies \dfrac{7}{11} + \dfrac{3\sqrt{3}}{11} - \dfrac{\sqrt{5}}{11} - \dfrac{2\sqrt{15}}{11} = P + \sqrt{3}Q + \sqrt{5}R + \sqrt{15}S}$

Comparing both sides, We can notice that :

$\mathsf{\implies P = \dfrac{7}{11}}$