Question

answer my question fast​

Answer 1

SOLUTION

Given,

$= > x = \frac{1}{2 - \sqrt{3} } \\ = > \frac{1}{2 - \sqrt{3} } \times \frac{(2 + \sqrt{3)} }{(2 + \sqrt{3} )} \\ = > \frac{(2 + \sqrt{3} }{(4 - 3)} \\ = > 2 + \sqrt{3}$

Now,

${x}^{3} - {2x}^{2} - 7x + 5 \\ = > (2 + \sqrt{3} ) { }^{3} - 2(2 + \sqrt{3} ) {}^{2} - 7(2 + \sqrt{3} ) + 5 \\ \\ = > 2 {}^{3} + ( \sqrt{3} ) {}^{3} + 3(4)( \sqrt{3} ) + 3(2)(3) - 2(4 + 3 + 4 \sqrt{3} ) - 14 - 7 \sqrt{3} + 5 \\ \\ = > 8 + 3 \sqrt{3} + 12 \sqrt{3} + 18 - 8 - 6 - 8 \sqrt{3} - 14 - 7 \sqrt{3} + 5 \\ \\ = > 8 + 18 - 8 - 6 - 14 + 5 + 3 \sqrt{3} + 12 \sqrt{3} - 8 \sqrt{3} - 7 \sqrt{ 3} \\ \\ = > 3 + 15 \sqrt{3} - 15 \sqrt{3} \\ \\ = > 3$

HOPE it helps ✔️

Answer 2

$12(1+\sqrt{3} )$

Step-by-step explanation:

$x=\frac{1}{2-\sqrt{3}} \times\frac{2+\sqrt{3} }{2+\sqrt{3}}\\\Rightarrow2+\sqrt{3}$

$p(x)=x^3-2x^2-7x+5\\p(2+\sqrt{3})^3-2(2+\sqrt{3})^2-7(2+\sqrt{3})+5\\ \Rightarrow2^3+(\sqrt{3)}^3+3\times2\times\sqrt{3}(2+\sqrt{3})-2\times2^2+(\sqrt3}^2)+2\times2\times\sqrt3-7(2+\sqrt3)+5\\\\\Rightarrow8+3\sqrt3+6\sqrt3(2+\sqrt3)-2\times4+3+4\sqrt3-14-7\sqrt3+5\\\Rightarrow8+3\sqrt3+18+12\sqrt3-8+3+4\sqrt3-14-7\sqrt3+5\\\Rightarrow3\sqrt3+12\sqrt3+4\sqrt3-7\sqrt3+12\\\Rightarrow12\sqrt3+72\\\Rightarrow12+12\sqrt3\\\Rightarrow12(1+\sqrt3)$