Question

if x=2-√3, find the value of (x+1/x)^3+ 2(x+1/x)^2+ (x+1/x)-100

Answer 1

1/x=1/2-√3;

2+√3/(2)^2-(√3)^2;

2+√3/1;

x+1/x=2-√3+2+√3=4

So value of (x+1/x)^3+ 2(x+1/x)^2+ (x+1/x)-100 is (4)^3+ 2(4)^2+ (4)-100

64+32-96

96-96=0

Answer 2

$\huge\underline\textsf{Answer:}$

Given: x=2-$\sf\sqrt{3}$, then

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

•°• x+$\sf\frac{1}{x}$=(2-√3)+(2+√3)=4.

Now, x+$\sf\frac{1}{x}$³+2(x+$\sf\frac{1}{x)}$²+(x+$\sf\frac{1}{x)}$-100

= (4)³+2(4)²+4-100

= 64+32+4-100

=100-100

=0.