Question

If x=1/2-√3,find the value of x^3-2x^2-7x+5​

Answer 1

Given :

$\longrightarrow \sf x = \dfrac{1}{2 - \sqrt{3} }$

To find :

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

Answer :

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

Rationalising the denominator :

Rationalising factor = 2 + √3

$\implies \sf \dfrac{1}{2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 + \sqrt{3} }$

$\implies \sf \dfrac{1(2 + \sqrt{3}) }{(2 - \sqrt{3})(2 + \sqrt{3} ) }$

The identity (a-b)(a+b) = a² - b² is formed in the denominator.

$\implies \sf \dfrac{2 + \sqrt{3} }{ {2}^{2} - { (\sqrt{3}) }^{2} }$

$\implies \sf \dfrac{2 + \sqrt{3} }{4 - 3}$

$\implies \sf \dfrac{2 + \sqrt{3} }{1}$

$\implies \sf x = 2 + \sqrt{3}$

x³ :

Applying (a+b)³ = a³ + 3ab(a+b) + b³ identity,

$\implies \sf (2 + \sqrt{3})^{3}$

$\implies \sf {2}^{3} + 3(2)( \sqrt{3})(2 + \sqrt{3} ) + ( \sqrt{3})^{3}$

$\implies \sf 8 + 6 \sqrt{3} (2 + \sqrt{3} ) + 3 \sqrt{3}$

$\implies \sf 8 + 12 \sqrt{3} + 18 + 3 \sqrt{3}$

$\implies \sf 26 + 15 \sqrt{3}$

2x² :

Applying (a+b)² = a² + 2ab + b² identity,

$\implies \sf 2(2 + \sqrt{3} )^{2}$

$\implies \sf 2\{ {2}^{2} + 2(2)( \sqrt{3} ) + ( \sqrt{3} )^{2} \}$

$\implies \sf 2\{4 + 4\sqrt{3} + 3 \}$

$\implies \sf 2 \{7 + 4 \sqrt{3} \}$

$\implies \sf 14 + 8 \sqrt{3}$

7x :

$\implies \sf 7(2 + \sqrt{3} )$

$\implies \sf 14 + 7 \sqrt{3}$

x³ - 2x² - 7x + 5

Substituting all the values,

$\implies \sf (26 + 15 \sqrt{3} ) - (14 + 8 \sqrt{3} ) - (14 + 7 \sqrt{3} ) + 5$

$\implies \sf 26 + 15 \sqrt{3} - 14 - 8 \sqrt{3} - 14 - 7 \sqrt{3} + 5$

$\implies \sf (26 - 14 - 14 + 5) + (15 \sqrt{3} - 8 \sqrt{3} - 7 \sqrt{3} )$

$\implies \sf (3) + (0)$

$= \sf 3$

Therefore, x³ - 2x² - 7x + 5 = 3

Answer 2

Step-by-step explanation:

Given :

$\longrightarrow \sf x = \dfrac{1}{2 - \sqrt{3} }$

To find :

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

Answer :

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

Rationalising the denominator :

Rationalising factor = 2 + √3

$\implies \sf \dfrac{1}{2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 + \sqrt{3} }$

$\implies \sf \dfrac{1(2 + \sqrt{3}) }{(2 - \sqrt{3})(2 + \sqrt{3} ) }$

The identity (a-b)(a+b) = a² - b² is formed in the denominator.

$\implies \sf \dfrac{2 + \sqrt{3} }{ {2}^{2} - { (\sqrt{3}) }^{2} }$

$\implies \sf \dfrac{2 + \sqrt{3} }{4 - 3}$

$\implies \sf \dfrac{2 + \sqrt{3} }{1}$

$\implies \sf x = 2 + \sqrt{3}$

x³ :

Applying (a+b)³ = a³ + 3ab(a+b) + b³ identity,

$\implies \sf (2 + \sqrt{3})^{3}$

$\implies \sf {2}^{3} + 3(2)( \sqrt{3})(2 + \sqrt{3} ) + ( \sqrt{3})^{3}$

$\implies \sf 8 + 6 \sqrt{3} (2 + \sqrt{3} ) + 3 \sqrt{3}$

$\implies \sf 8 + 12 \sqrt{3} + 18 + 3 \sqrt{3}$

$\implies \sf 26 + 15 \sqrt{3}$

2x² :

Applying (a+b)² = a² + 2ab + b² identity,

$\implies \sf 2(2 + \sqrt{3} )^{2}$

$\implies \sf 2\{ {2}^{2} + 2(2)( \sqrt{3} ) + ( \sqrt{3} )^{2} \}$

$\implies \sf 2\{4 + 4\sqrt{3} + 3 \}$

$\implies \sf 2 \{7 + 4 \sqrt{3} \}$

$\implies \sf 14 + 8 \sqrt{3}$

7x :

$\implies \sf 7(2 + \sqrt{3} )$

$\implies \sf 14 + 7 \sqrt{3}$

x³ - 2x² - 7x + 5

Substituting all the values,

$\implies \sf (26 + 15 \sqrt{3} ) - (14 + 8 \sqrt{3} ) - (14 + 7 \sqrt{3} ) + 5$

$\implies \sf 26 + 15 \sqrt{3} - 14 - 8 \sqrt{3} - 14 - 7 \sqrt{3} + 5$

$\implies \sf (26 - 14 - 14 + 5) + (15 \sqrt{3} - 8 \sqrt{3} - 7 \sqrt{3} )$

$\implies \sf (3) + (0)$

$= \sf 3$

Therefore, x³ - 2x² - 7x + 5 = 3