Question

$x=\frac{1}{2+\sqrt{3} }$ Find Value of $2x^{3} - 7x^{2} -2x+1$

Please give proper answer with explaination

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:x = \dfrac{1}{2 + \sqrt{3} }$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:Value \: of \: {2x}^{3} - {7x}^{2} - 2x + 1$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x = \dfrac{1}{2 + \sqrt{3} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:x = \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$\rm :\longmapsto\:x = \dfrac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} }$

$\rm :\longmapsto\:x = \dfrac{2 - \sqrt{3} }{4 - 3}$

$\rm :\longmapsto\:x = 2 - \sqrt{3}$

$\rm :\longmapsto\:x - 2 = - \sqrt{3}$

On squaring both sides, we get

$\rm :\longmapsto\: {(x - 2)}^{2} = {( - \sqrt{3}) }^{2}$

$\rm :\longmapsto\: {x}^{2} + 4 - 4x = 3$

$\rm :\longmapsto\: {x}^{2} + 4 - 4x - 3 = 0$

$\rm :\longmapsto\: {x}^{2} - 4x + 1 = 0 - - - (1)$

Now, Consider,

$\rm :\longmapsto \: {2x}^{3} - {7x}^{2} - 2x + 1$

can be rewritten as

$\rm \: = \: \: {2x}^{3} + ( \red{ \rm - {8x}^{2} + {x}^{2}}) + ( \red{ \rm \: - 4x +2x}) + 1$

$\rm : = \: \: \: {2x}^{3} - {8x}^{2} + {x}^{2} - 4x + 2x + 1$

$\rm : = \: \: \: ( {2x}^{3} - {8x}^{2} + 2x) + ({x}^{2}-4x+1)$

$\rm \: = \: \: 2x( {x}^{2} - 4x + 1) + 0 \: \: \: \{ \: using \: (1) \: \}$

$\rm \: = \: \: 2x \times 0 \: \: \: \: \{ \: using \: (1) \: \}$

$\bf\implies \:\:Value \: of \: {2x}^{3} - {7x}^{2} - 2x + 1 = 0$

Additional Information :-

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)