Question

if x=1/2-sq root 3,find the value of x3-2x2-7x+5

Answer 1

$x=\frac{1}{2-\sqrt{3}}$
Multiply the numerator and denominator by the conjugate of the denominator
$x=\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$
$\implies\,x^2=(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}$

Now consider the given expression
$x^3-2x^2-7x+5\\=x^3-7x-2x^2+5\\=x(x^2-7)-2x^2+5\\=(2+\sqrt{3})(7+4\sqrt{3}-7)-2(7+4\sqrt{3})+5\\=(2+\sqrt{3})(4\sqrt{3})-14-8\sqrt{3}+5\\=8\sqrt{3}+12-14-8\sqrt{3}+5\\=\boxed{3}$

Answer 2

we rationalize the denominator to find x.
$x = \frac{1}{2-sqrt3}=\frac{2+\sqrt3}{(2+\sqrt3)(2-\sqrt3)}\\\\=\frac{2+\sqrt3}{2^2-3}=2+\sqrt3$

since, the expression of x is simple, we find x² and x³  by simple multiplication.  If the expression for x were complicated, then we could find some other simple means.

$x^2=(2+\sqrt3)^2=4+3+4\sqrt3=7+4\sqrt3\\\\x^3=x*x^2=(2+\sqrt3)(7+4\sqrt3)= 14+4*3+15\sqrt3$


Now we calculate the value of the polynomial in x:
$x^3 - 2 x^2 - 7 x + 5\\\\=26 + 15 \sqrt3 - 2(7+4\sqrt3) - 7 (2+\sqrt3)+5\\\\=26-14-14+ (15-8-7) \sqrt3+5\\\\=3$