Question

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Answer 1

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Answer 2

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

On rationalizing, we get

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

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

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

$= > x = 2 + \sqrt{3}$

Now,

Given : x^3 - 2x^2 - 7x + 5.

⇒ (2 + √3)^3 - 2(2 + √3)^2 - 7(2 + √3) + 5

⇒ 8 + 3√3 + 12√3 + 18 - 2(4 + 3 + 4√3) - 14 - 7√3 + 5

⇒ 8 + 3√3 + 12√3 + 18 - 8 - 6 - 8√3 - 14 - 7√3 + 5

⇒ 3.

Therefore, the value of x^3 - 2x^2 - 7x + 5 = 3.

Hope this helps