Question

If $x=\frac{1}{2-\sqrt{3}}$, find the value of $x^3-2x^2-7x+5=m$.Find $m$

Answer 1

Solution:-

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

Multiplying the numerator and denominator by the conjugate of the denominator, we have

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

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

Now consider the given expression

$x^3-2x^2-7x+5=x^3-7x-2x^2+5$

$\Rightarrow =x(x^2-7)-2x^2+5$

$\Rightarrow =(2+\sqrt{3})(7+4\sqrt{3}-7)-2(7+4\sqrt{3})+5$

$\Rightarrow =(2+\sqrt{3})\left(4\sqrt{3}\right)-14-8\sqrt{3}+5$

$\Rightarrow =8\sqrt{3}+12-14-8\sqrt{3}+5$

$\Rightarrow =3$

$x^3-2x^2-7x+5=m\left\{Given\right\}$

$\therefore m=3$