Question

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

Answer 1

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Answer 2

Question -

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

Solution -

Given,

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

By rationalising the denominator we will get -

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

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

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

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

$x = 2 + \sqrt{3}$

Now,

$= {x}^{2} - 4x + 1$

Putting the value of x we will get -

$= {(2 + \sqrt{3}) }^{2} - 4(2 + \sqrt{3} ) + 1$

$= {(2)}^{2} + ({ \sqrt{3} )}^{2} + 2 \times 2 \times \sqrt{3} - 8 - 4 \sqrt{3} + 1$

$= 4 + 3 + 4 \sqrt{3} - 8 - 4 \sqrt{3} + 1$

$= 4 + 3 - 8 + 1$

$= 8 - 8$

$= 0$

Therefore the value of x²-4x+1 is 0