Question

if x=1/2-root 3,find the value of x2-4x+1=0​

Answer 1

Answer:

x² - 4x + 1 = 16 + 8√3

Step-by-step explanation:

Given x = 1/2-√3

x = 1/(2 - √3) × (2 + √3)/(2 + √3)

x = 2 + √3/2 - 3

x = -(2 + √3)

x² = 4 + 3 + 4√3

x² = 7 + 4√3

we have to find

x² - 4x + 1

=> 7 + 4√3 - 4(-2-√3) + 1

=> 7 + 4√3 + 8 + 4√3 + 1

=> 15 + 8√3 + 1

=> 16 + 8√3

=> x² - 4x + 1 = 16 + 8√3

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Answer 2

The value of x² - 4x + 1 = 0

Correct question :

$\displaystyle \sf If \: x = \frac{1}{2 - \sqrt{3} } \: find \: the \: value \: of \: {x}^{2} - 4x + 1$

Given :

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

To find :

The value of x² - 4x + 1

Solution :

Step 1 of 2 :

Rationalize the denominator of the value of x

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

$\displaystyle \sf{ \implies }x = \frac{2 + \sqrt{3} }{(2 + \sqrt{3})(2 - \sqrt{3} ) }$

$\displaystyle \sf{ \implies }x = \frac{2 + \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} }$

$\displaystyle \sf{ \implies }x = \frac{2 + \sqrt{3} }{4 - 3}$

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

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

Step 2 of 2 :

Find the value of x² - 4x + 1

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

$\displaystyle \sf{ \implies }x - 2 = \sqrt{3}$

Squaring both sides we get

$\displaystyle \sf{ }{(x - 2)}^{2} = {(\sqrt{3})}^{2}$

$\displaystyle \sf{ \implies } {x}^{2} - 4x + 4 = 3$

$\displaystyle \sf{ \implies } {x}^{2} - 4x + 4 - 3 = 0$

$\displaystyle \sf{ \implies } {x}^{2} - 4x + 1 = 0$

Hence the value of x² - 4x + 1 = 0

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