Question

If $x=4+\sqrt{15}$ , find the value of $a)x-\dfrac{1}{x}$ and $b){{x}^{2}}+\dfrac{1}{{{x}^{2}}}$.

Answer 1

Hint: To solve this question we need to have the concept of Binomial expansion and rationalization. In the first part of the question we will be find the value of \[\dfrac{1}{x}\]and then subtract it from$x$. the second question we will be applying the formula for a plus b whole square and then finding the value for a square + b square respectively.

Complete step by step answer:

The question ask us to find the value of $x-\dfrac{1}{x}$ and ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ when value for the “x” is given to us which is a complex number, which is $4+\sqrt{15}$. 

a) We will be first working upon the solution of $x-\dfrac{1}{x}$. Currently, we know the value of \[x-\dfrac{1}{x}\] which is $4+\sqrt{15}$. We will find the reciprocal of $x$ number, which means \[\dfrac{1}{x}\] . 

Since we know that the number need to be nationalized so as to remove the complex part of the number from the denominator on rationalizing the above fraction we get \[\dfrac{1}{x}\], which is 

$\Rightarrow \dfrac{1}{x}=\dfrac{1}{4+\sqrt{15}}$

On rationalising we get:

\[\Rightarrow \dfrac{1}{x}=\dfrac{1}{4+\sqrt{15}}\times \dfrac{\left( 4-\sqrt{15} \right)}{\left( 4-\sqrt{15} \right)}\]

\[\Rightarrow \dfrac{1}{x}=\dfrac{1\left( 4-\sqrt{15} \right)}{\left( 4+\sqrt{15} \right)\left( 4-\sqrt{15} \right)}\]

On analysing the above expression we see that we can apply the formula$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ . On applying the same in the denominator of the fraction we get:

\[\Rightarrow \dfrac{1}{x}=\dfrac{1\left( 4-\sqrt{15} \right)}{{{4}^{2}}-{{\left( \sqrt{15} \right)}^{2}}}\]

\[\Rightarrow \dfrac{1}{x}=\dfrac{4-\sqrt{15}}{16-15}\]

On calculating further we get:

\[\Rightarrow \dfrac{1}{x}=\dfrac{4-\sqrt{15}}{1}\]

So, the value of \[\dfrac{1}{x}\] becomes \[4-\sqrt{15}\].

Now, on calculating the expression $x-\dfrac{1}{x}$ we get:

$\Rightarrow x-\dfrac{1}{x}=4+\sqrt{15}-4+\sqrt{15}$ 

The term $\sqrt{15}$ get cancelled so the value becomes:

$\Rightarrow x-\dfrac{1}{x}=2\sqrt{15}$

  

b) Now we will find the solution of the second part of the question, which is ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ . To solve the problem we will apply the formula${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ . We will substitute the value of \[x\] and \[\dfrac{1}{x}\] respectively. On doing this we get:

$\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2\times x\times \dfrac{1}{x}$ 

$\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2$

We know the value of L.H.S from the section a) of the question. So finding the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$is an easy task.

$\Rightarrow {{\left( 2\sqrt{15} \right)}^{2}}={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2$

$\Rightarrow 60={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2$

Adding the number $2$ to both sides of the equation. On doing this we get:

$\Rightarrow 60+2={{x}^{2}}+\dfrac{1}{{{x}^{2}}}$

$\Rightarrow 62={{x}^{2}}+\dfrac{1}{{{x}^{2}}}$

Note: Remember that the imaginary part in the denominator needs to be changed as the denominator cannot be a complex number. It needs to be changed to a real one so that we will be multiplying both numerator and denominator by the conjugate of the denominator. Conjugate of $a+\sqrt{b}$ is called $a-\sqrt{b}$ , the conjugate of a complex number refers to the same real and imaginary number where the sign of the imaginary number is opposite. Example: Conjugate of $-3-\sqrt{5}$ is $-3+\sqrt{5}$.