Question

If \[x=\dfrac{1}{2}\left( \sqrt{7}+\left( \dfrac{1}{\sqrt{7}} \right) \right)\], then \[\dfrac{\left( \sqrt{{{x}^{2}}-1} \right)}{\left[ x-\sqrt{{{x}^{2}}-1} \right]}\]\[=\]
\[\begin{align}
  & a.1 \\
 & b.2 \\
 & c.3 \\
 & d.4 \\
\end{align}\]

Answer 1

Hint: In order to find the value of \[\dfrac{\left( \sqrt{{{x}^{2}}-1} \right)}{\left[ x-\sqrt{{{x}^{2}}-1} \right]}\], when \[x=\dfrac{1}{2}\left( \sqrt{7}+\left( \dfrac{1}{\sqrt{7}} \right) \right)\], then we will be squaring the given \[x\] value on both sides and obtain the value of \[x\] in a simplified manner. Then after obtaining the value of \[x\], we will be substituting this value in the expression given in order to obtain its value.

Complete step by step answer:
Now let us learn more about the quadratic equations. A quadratic equation is nothing but the equation whose degree is two. The general form of a quadratic equation is \[a{{x}^{2}}+bx+c=0\] where \[a\ne 0\]. The values that satisfy the equation for the variable, such values are called the roots of the quadratic equation. A quadratic equation will definitely have two roots. The discriminant of a quadratic equation defines the nature of the roots of the equation. The discriminant is nothing but the term \[{{b}^{2}}-4ac\]. The roots of a quadratic equation can be found by the method factorization or by using the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Now let us start finding the value of\[\dfrac{\left( \sqrt{{{x}^{2}}-1} \right)}{\left[ x-\sqrt{{{x}^{2}}-1} \right]}\].
So we are given with the value of \[x=\dfrac{1}{2}\left( \sqrt{7}+\left( \dfrac{1}{\sqrt{7}} \right) \right)\].
Now let us solve this by squaring it on both sides using \[{{\left( a+b \right)}^{2}}=a^2+b^2+2ab\], Upon doing so, we obtain
\[\begin{align}
  & x=\dfrac{1}{2}\left( \sqrt{7}+\left( \dfrac{1}{\sqrt{7}} \right) \right) \\
 & {{x}^{2}}=\dfrac{1}{4}\left( 7+\dfrac{1}{7}+2 \right) \\
 & \Rightarrow 4{{x}^{2}}=7+\dfrac{1}{7}+2 \\
\end{align}\]
Now let us subtract \[4\] on both sides, then we get
\[\begin{align}
  & \Rightarrow 4{{x}^{2}}-4=7+\dfrac{1}{7}+2-4 \\
 & \Rightarrow 4{{x}^{2}}-4=7+\dfrac{1}{7}-2 \\
\end{align}\]
Now we will be expressing the RHS in terms of the identity \[{{\left( a-b \right)}^{2}}=a^2+b^2-2ab\],
\[\begin{align}
  & \Rightarrow 4{{x}^{2}}-4=7+\dfrac{1}{7}-2 \\
 & \Rightarrow 4\left( {{x}^{2}}-1 \right)={{\left( 7-\dfrac{1}{7} \right)}^{2}} \\
\end{align}\]
Upon solving this further, we get
\[\begin{align}
  & \Rightarrow \left( {{x}^{2}}-1 \right)={{\left( \dfrac{1}{2}\left( \sqrt{7}-\dfrac{1}{\sqrt{7}} \right) \right)}^{2}} \\
 & \Rightarrow x-\sqrt{{{x}^{2}}-1}=\dfrac{1}{2}\left( \sqrt{7}+\dfrac{1}{\sqrt{7}} \right)-\dfrac{1}{2}\left( \sqrt{7}-\dfrac{1}{\sqrt{7}} \right) \\
 & \Rightarrow \left( \dfrac{1}{2} \right)\left( \sqrt{7}+\dfrac{1}{\sqrt{7}} -\sqrt{7}+\dfrac{1}{\sqrt{7}}\right) \\
 & \Rightarrow \dfrac{1}{\sqrt{7}} \\
\end{align}\]
Now let us substitute the obtained values in the expression \[\dfrac{\left( \sqrt{{{x}^{2}}-1} \right)}{\left[ x-\sqrt{{{x}^{2}}-1} \right]}\]
\[\begin{align}
  & \dfrac{\left( \sqrt{{{x}^{2}}-1} \right)}{\left[ x-\sqrt{{{x}^{2}}-1} \right]}=\dfrac{\dfrac{1}{2}\left( \sqrt{7}-\dfrac{1}{\sqrt{7}} \right)}{\dfrac{1}{\sqrt{7}}} \\
 & \Rightarrow \dfrac{1}{2}\left( 7-1 \right)=\dfrac{1}{2}\left( 6 \right)=3 \\
\end{align}\]

So, the correct answer is “Option c”.

Note: We must always note that we must choose the appropriate methods to solve such problems. We must try to expand or express them in the form of identities for easy simplification of the equations. When two expressions are given, we must choose the simpler one to solve.