Question

$If \: x=\frac{1}{2 - \sqrt{3} }, then \: find \: the \: value \: of \: x³ - 2 {x}^{2} - 7x + 5 \: Ifx=2−3​1​, \: then \: find \: the \: value \: of \: x³−2x2−7x+5$
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Answer 1

Given that :

$\sf : \; \implies x = \dfrac{1}{2-\sqrt{3}}$

Need to find : x³ - 2x² - 7x + 5

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❍ To calculate the value of given expression we have to rationalize the denominator of the given value of x first.

$\sf : \; \implies x = \dfrac{1}{2-\sqrt{3}}$

$\sf : \; \implies x = \dfrac{1}{2-\sqrt{3}} \times \dfrac{2 +\sqrt{3}}{2 + \sqrt{3}}$

$\sf : \; \implies x = \dfrac{2+\sqrt{3}}{(2-\sqrt{3})(2 + \sqrt{3})}$

$\sf : \; \implies x = \dfrac{2+\sqrt{3}}{(2)^{2}- (\sqrt{3})^{2}}$

$\sf : \; \implies x = \dfrac{2+\sqrt{3}}{4-3}$

$\sf : \; \implies x = \dfrac{2+\sqrt{3}}{1}$

$\sf \bullet \;\; \twoheadrightarrow x = 2 + \sqrt{3}$

T H E R E F O R E,

$\sf : \; \implies x^{3} - 7x - 2x^{2} + 5$

$\sf : \; \implies x(x^{2}-7) - 2x^{2} + 5$

$\sf : \; \implies 2 + \sqrt{3} ( [\; 2 + \sqrt{3} \;]^{2}-7) - 2 \times ( 2 +\sqrt{3} )^{2} + 5$

$\sf : \; \implies 2 + \sqrt{3} ( [ \;4+ 4\sqrt{3} + 3 \; ]-7) - 2 \times ( 4+ 4\sqrt{3} + 3) + 5$

$\sf : \; \implies 2 + \sqrt{3} ( 4\sqrt{3} + 7 -7) - 2 \times( 4\sqrt{3} +7) + 5$

$\sf : \; \implies 2 + \sqrt{3} ( 4\sqrt{3} ) - 8\sqrt{3} - 14 + 5$

$\sf : \; \implies 8\sqrt{3} + 12 - 8\sqrt{3} - 14 + 5$

$\sf : \; \implies 8\sqrt{3} - 8\sqrt{3} - 14 + 5 + 12$

$\sf : \; \implies - 14 + 5 + 12$

$\sf : \; \implies 17 - 14$

$\boxed{\bf{ \bigstar \;\; \leadsto 3}}$

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• Henceforth, the value of the given expression is 3.

Answer 2

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