Question

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

Answer 1

Hint: Here we are going to apply the basic formulas of conjugates. Such as, the conjugate of \[\left( a+\sqrt{b} \right)\]is \[\left( a-\sqrt{b} \right)\].To avoid the complexity of calculations.
Formulas used:
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\]

Complete step-by-step answer:
\[x=\dfrac{1}{2-\sqrt{3}}\] \[\begin{align}
  & =\dfrac{1}{2-\sqrt{3}}.\dfrac{2+\sqrt{3}}{2+\sqrt{3}}\text{ }\left[ Multiplyin\text{g both numerator }\!\!\And\!\!\text{ denominator by conjugate of denominator} \right] \\
 & =\dfrac{2+\sqrt{3}}{{{\left( 2 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}} \\
 & =\dfrac{2+\sqrt{3}}{4-3} \\
 & =2+\sqrt{3} \\
\end{align}\]
We have to calculate \[{{x}^{2}}\]separately for the ease of the calculation.
\[{{x}^{2}}={{\left( 2+\sqrt{3} \right)}^{\begin{smallmatrix}
 2 \\

\end{smallmatrix}}}\]
\[\begin{align}
  & =4+4\sqrt{3}+3\text{ }\left[ We\text{ know that,}{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right] \\
 & =7+4\sqrt{3} \\
\end{align}\]
Now if we consider the given expression-
\[\begin{align}
  & {{x}^{3}}-2{{x}^{2}}-7x+5 \\
 & ={{x}^{3}}-7x-2{{x}^{2}}+5 \\
 & =x\left( {{x}^{2}}-7 \right)-2{{x}^{2}}+5 \\
 & =\left( 2+\sqrt{3} \right)\left( 7+4\sqrt{3}-7 \right)-2\left( 7+4\sqrt{3} \right)+5\text{ }\left[ \text{Now, replacing the value of x }\!\!\And\!\!\text{ }{{\text{x}}^{2}} \right] \\
 & =\left( 2+\sqrt{3} \right)4\sqrt{3}-14-8\sqrt{3}+5 \\
 & =8\sqrt{3}+12-14-8\sqrt{3}+5 \\
 & =3 \\
\end{align}\]
Hence
    For, \[x=\dfrac{1}{2-\sqrt{3}}\] the value of \[{{x}^{3}}-2{{x}^{2}}-7x+5\]will be \[3\].

Note: What is a complex conjugate?
Each of two complex numbers having their real parts identical and their imaginary parts of equal magnitude but opposite in sign are called the conjugate multiple of one another.