Question

if a=√5+1/√5-1 and b=√5-1/√5+1 find the value of 3a^2+5ab+3b^2/3a^2-5ab+3b^2​

Answer 1

Answer:

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Answer 2

$\Huge \bf{Given:-}$

$\bf a=\dfrac{\sqrt{5+1}}{\sqrt{5-1}}$

$\bf b=\dfrac{\sqrt{5-1}}{\sqrt{5+1}}$

$\Huge \bf{To\:Find:-}$

$\bf Value \:of\: \dfrac{3a^2+5ab+3b^2}{3a^2-5ab+3b^2}$

$\Huge \bf{Solution:-}$

$\bf Rationalizing\:the \: denominator$

$\bf a=\dfrac{\sqrt{5+1}}{\sqrt{5-1}}$

$\bf \implies \dfrac{\sqrt{5+1}}{\sqrt{5-1}}×\dfrac{\sqrt{5+1}}{\sqrt{5+1}}$

$\bf \implies \dfrac{\sqrt{5+1}^2}{5-1}$

$\bf \implies \dfrac{(\sqrt{5})^2+(\sqrt{1})^2+(2×\sqrt{5}×\sqrt{1})}{4}$

$\bf \implies \dfrac{5+1+(2\sqrt{5})}{4}$

$\bf \implies \dfrac{6+(2\sqrt{5})}{4}$

$\bf \implies \dfrac{3+(\sqrt{5})}{2}$

$\bf b=\dfrac{\sqrt{5-1}}{\sqrt{5+1}}$

$\bf \implies \dfrac{\sqrt{5-1}}{\sqrt{5+1}}×\dfrac{\sqrt{5-1}}{\sqrt{5-1}}$

$\bf \implies \dfrac{\sqrt{5-1}^2}{5-1}$

$\bf \implies \dfrac{(\sqrt{5})^2+(\sqrt{1})^2-(2×\sqrt{5}×\sqrt{1})}{4}$

$\bf \implies \dfrac{5+1-(2\sqrt{5})}{4}$

$\bf \implies \dfrac{6-(2\sqrt{5})}{4}$

$\bf \implies \dfrac{3-(\sqrt{5})}{2}$

$\bf putting \:Values \: in \dfrac{3a^2+5ab+3b^2}{3a^2-5ab+3b^2}$

$\begin{gathered}\bf \implies \dfrac{3(\dfrac{3+(\sqrt{5}}{2})^2+5(\dfrac{3+\sqrt{5}}{2})(\dfrac{3-\sqrt{5}}{2})+3(\dfrac{3-\sqrt{5}}{2})^2}{{3(\dfrac{3+\sqrt{5}}{2})^2-5(\dfrac{3+\sqrt{5}}{2})(\dfrac{3-\sqrt{5}}{2})+3(\dfrac{3-\sqrt{5}}{2})^2}\end{gathered}$