Question

If X = 3 + 2root2 then find the value of rootx -1/rootx​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$\mathsf{Given :\;x = 3 + 2\sqrt{2}}$

$\mathsf{\implies x = 1 + 2 + 2\sqrt{2}}$

$\mathsf{\implies x = (1)^2 + \big(\sqrt{2}\big)^2 + 2\sqrt{2}}}$

Above expression is in the form :

★  a² + b² + 2ab which is equal to (a + b)²

$\mathsf{Where : a = 1\;and\;b = \sqrt{2}}$

$\mathsf{\implies (1)^2 + \big(\sqrt{2}\big)^2 + 2\sqrt{2} = \big(1 + \sqrt{2}\big)^2}$

$\mathsf{\implies x = \big(1 + \sqrt{2}\big)^2}$

$\mathsf{\implies \sqrt{x} = \sqrt{\big(1 + \sqrt{2}\big)^2}}$

$\implies \boxed{\mathsf{\sqrt{x} = 1 + \sqrt{2}}}$

$\mathsf{\implies \dfrac{1}{\sqrt{x}} = \dfrac{1}{1 + \sqrt{2}}}$

$\mathsf{Multiplying\;numerator\;and\;denominator\;with\;1 - \sqrt{2},\;We\;get :}$

$\mathsf{\implies \dfrac{1}{\sqrt{x}} = \dfrac{1 - \sqrt{2}}{(1 + \sqrt{2})(1 - \sqrt{2})}}$

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

$\mathsf{\implies \dfrac{1}{\sqrt{x}} = \dfrac{1 - \sqrt{2}}{(1)^2 - (\sqrt{2})^2}}$

$\mathsf{\implies \dfrac{1}{\sqrt{x}} = \dfrac{1 - \sqrt{2}}{1 - 2}}$

$\mathsf{\implies \dfrac{1}{\sqrt{x}} = \dfrac{1 - \sqrt{2}}{-1}}$

$\implies\boxed{\mathsf{\dfrac{1}{\sqrt{x}} = \sqrt{2} - 1}}$

$\mathsf{\implies \sqrt{x} - \dfrac{1}{\sqrt{x}} = 1 + \sqrt{2} - (\sqrt{2} - 1)}$

$\mathsf{\implies \sqrt{x} - \dfrac{1}{\sqrt{x}} = 1 + \sqrt{2} - \sqrt{2} + 1}$

$\mathsf{\implies \sqrt{x} - \dfrac{1}{\sqrt{x}} = 2}$