Question

If x=1/2-√3,find x^3-2x^2-7x+4.

Plz....Plz....Please answer it....Plz....

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = \frac{1}{2 - \sqrt{3} }$

On rationalizing the denominator we get :

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

Using the identity :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

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

Putting the value of x we get,

${x}^{3} - {2x}^{2} - 7x + 4 \\ \\ = {(2 + \sqrt{3} )}^{3} - 2 {(2 + \sqrt{3}) }^{2} - 7(2 + \sqrt{3} ) + 4$

Using the identities :

${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \\ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

$= ( {(2)}^{3} + {( \sqrt{3}) }^{3} + 3(2)( \sqrt{3} )(2 + \sqrt{3} )) - 2( {(2)}^{2} + {( \sqrt{3}) }^{2} + 2(2)( \sqrt{3} )) - 7 \times 2 - 7 \times \sqrt{3} + 4 \\ \\ = (8 + 3 \sqrt{3} + 6 \sqrt{3} (2 + \sqrt{3} )) - 2(4 + 3 + 4 \sqrt{3} ) - 14 - 7 \sqrt{3} + 4 \\ \\ = 8 + 3 \sqrt{3} + 12 \sqrt{3} + 18 - 8 - 6 - 8 \sqrt{3} - 14 - 7 \sqrt{3} + 4 \\ \\ = 8 + 18 + 4 - 8 - 6 - 14 + 3 \sqrt{3} + 12 \sqrt{3} - 8 \sqrt{3} - 7 \sqrt{3} \\ \\ = 30 - 28 + 15 \sqrt{3} - 15 \sqrt{3} \\ \\ = 2$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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