Question

If $x=2-\sqrt{3}$ , then find the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ .
(a) 14
(b) -14
(c) 2
(d) -2

Answer 1

Hint: Use the formula ${{\left( a-b \right)}^{2}}+2ab={{a}^{2}}+{{b}^{2}}$ , for simplification of the expression ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ followed by substituting the value of x to get the required answer. You might need to use rationalisation to reach the result.

Complete step-by-step answer:
The expression given in the question is:
 ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$
We know that ${{\left( a-b \right)}^{2}}+2ab={{a}^{2}}+{{b}^{2}}$ . So, if we use this to simplify the right-hand side of our equation, we get
${{\left( x-\dfrac{1}{x} \right)}^{2}}+2x\times \dfrac{1}{x}$
$={{\left( x-\dfrac{1}{x} \right)}^{2}}+2$
Now if we substitute the value of x as $2-\sqrt{3}$ , we get
 ${{\left( 2-\sqrt{3}-\dfrac{1}{2-\sqrt{3}} \right)}^{2}}+2$
Now we would rationalise the denominator of the term with irrational denominator. On, doing so, we get
${{\left( 2-\sqrt{3}-\dfrac{\left( 2+\sqrt{3} \right)}{\left( 2-\sqrt{3} \right)\left( 2+\sqrt{3} \right)} \right)}^{2}}+2$
$={{\left( 2-\sqrt{3}-\dfrac{\left( 2+\sqrt{3} \right)}{{{2}^{2}}-3} \right)}^{2}}+2$
$={{\left( 2-\sqrt{3}-2-\sqrt{3} \right)}^{2}}+2$
$={{\left( -2\sqrt{3} \right)}^{2}}+2$
$=12+2=14$
Therefore, we can conclude that the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ is 14. Hence, the answer to the above question is option (a).

Note: We could have also solved the above question by changing the equation ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ into a biquadratic polynomial by taking ${{x}^{2}}$ as the LCM in first step, but that could have been lengthier and more complex to solve. Also, be careful about the signs in the formulas that we used and what is asked in the question. It is also necessary that you learn all the basic algebraic formulas and binomial expansions as they are often used in solving problems.