Question

x is equal to 1 upon 2 + root 3 find the value of 2 x cube minus 7 x square - 2 X + 1

Answer 1

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Answer 2

$\underline{\mathbf{Formula \: used : }}$

$\boxed{\mathbf{{(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b)}}$

$\boxed{\mathbf{{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}}$

$\boxed{\mathbf{(a - b)(a - b) = {a}^{2} - {b}^{2}}}$

$\underline{\underline{\huge\mathfrak{Solution}}}$

Given,

$x = \frac{1}{2 + \sqrt{3} }$

Rationalising the denominator we get,

$x = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ x = \frac{2 - \sqrt{3} }{(2 + \sqrt{3} )(2 - \sqrt{3}) } \\ \\ x = \frac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } \\ \\ x = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \bf x = 2 - \sqrt{3}$

Now, putting the value of x in the given equation we get :

$2 {x}^{3} - 7 {x}^{2} - 2x + 1 \\ \\ = 2 {(2 - \sqrt{3} )}^{3} - 7( {2 - \sqrt{3}) }^{2} - 2(2 - \sqrt{3} ) + 1 \\ \\ \bf putting \: the \: formula \: we \: get \\ \\ = 2( {(2)}^{3} - {( \sqrt{3}) }^{3} - 3(2)( \sqrt{3} )(2 - \sqrt{3} ) )\\ \: \: \: \: - 7( {(2)}^{2} + {( \sqrt{3} )}^{2} - 2(2)( \sqrt{3} ) \\ \: \: \: \: - 2(2) + 2( \sqrt{3} ) + 1 \\ \\ = 2(8 - 3 \sqrt{3} - 6 \sqrt{3} (2 - \sqrt{3} )) \\ - 7( 4+ 3 - 4 \sqrt{3} ) - 4 + 2 \sqrt{3} + 1 \\ \\ = 2(8 - 3 \sqrt{3} - 12 \sqrt{3} + 18) \\ - 7( 7 - 4 \sqrt{3} ) - 3 + 2 \sqrt{3} \\ \\ = 2(26 - 15 \sqrt{3} ) - 49 + 28 \sqrt{3} - 3 + 2 \sqrt{3} \\ \\ = 52 - 30 \sqrt{3} - 49 + 28 \sqrt{3} - 3 + 2 \sqrt{3} \\ \\ = 52 - 49 - 3 - 30 \sqrt{3} + 28 \sqrt{3} + 2 \sqrt{3} \\ \\ = 52 - 52 - 30 \sqrt{3} + 30 \sqrt{3}$

= $\boxed{\boxed{\mathbf{ 0}}}$