Question

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

Answer 1

Hint: In this problem, first we will rationalize the value of x to make denominator rational, that is after rationalization we get $x=2+\sqrt{3}$ then applying some manipulation on x, that is bringing 2 to the left side and squaring both sides, after that, we will get an equation in terms ${{x}^{2}}$ , further multiplying it with x, we get an equation in terms of ${{x}^{3}}$ then after substituting the value ${{x}^{2}}$and ${{x}^{3}}$in the expression ${{x}^{3}}-2{{x}^{2}}-7x+5$, we get the desired result.

Complete step by step answer:
Here we have,
 $x=\dfrac{1}{2-\sqrt{3}}$ , then rationalsing the denominator, we have
$x=\dfrac{1}{\left( 2-\sqrt{3} \right)}\times \dfrac{2+\sqrt{3}}{2+\sqrt{3}}$
After , simplifying we get
 $x=\dfrac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$
Now , bringing 2 to the left side and squaring both sides, we have
${{\left( x-2 \right)}^{2}}={{\left( \sqrt{3} \right)}^{2}}$
Simplifying further we get,
$\begin{align}
  & {{x}^{2}}-4x+4=3 \\
 & \Rightarrow {{x}^{2}}-4x=-1\,\,\,\,\,\,\,...........(i) \\
\end{align}$
Now , writing whole expression in terms of ${{x}^{2}}$ , we have
${{x}^{2}}=4x-1$ ……….(ii)
Now , multiplying equation (ii) with x, we have
${{x}^{3}}=4{{x}^{2}}-x$ ………(iii)
Now, we have
${{x}^{3}}-2{{x}^{2}}-7x+5$
Putting the value of ${{x}^{3}}$ in the above expression, from the equation (iii), we have
$\begin{align}
  & 4{{x}^{2}}-x-2{{x}^{2}}-7x+5 \\
 & =2{{x}^{2}}-8x+5 \\
\end{align}$
Now , taking 2 common from first two terms, we have
$2\left( {{x}^{2}}-4x \right)+5$
As , ${{x}^{2}}-4x=-1$ from equation (i), substituting this value in above expression, we get
$\begin{align}
  & 2(-1)+5 \\
 & =-2+5 \\
 & =3 \\
\end{align}$
$\therefore $ the anser is 3
Hence, when $x=\dfrac{1}{2-\sqrt{3}}$, then ${{x}^{3}}-2{{x}^{2}}-7x+5=3$

Note:
Always remember that if in any expression denominator is irrational make it rational by rationalization then proceed further. In this type of problem never try to directly substitute the value of x as the expression contains the cubic and square term, the calculation will become more complex and will be very hectic to solve. So, always try to apply some manipulation on x as solved above to make calculations easier.