Question

$if \: {a}^{2} + \frac{1}{ {a}^{2} } then \: find \: \: the \: value \: of \: a - \frac{1}{a}$
answer the question above ​

Answer 1

$\huge{\pm\sqrt{x-2}}$

$\huge\text{\underline{\underline{Note}}}$

*Check the equation again. I am assuming the first value as $x$.

$\huge\text{\underline{\underline{Review}}}$

$\Large\text{\{Polynomials\}}$

$\large\text{ \longrightarrow Polynomial identity}$

An equation that both the L.H.S and R.H.S are always equivalent is called an identity.

One of the important equations is the following identity.

$\bold{Perfect\ square\ identity}$

$\text{ \cdots\longrightarrow\boxed{(A+B)^{2}=A^{2}+2AB+B^{2}} }$

$\huge\text{\underline{\underline{Question}}}$

If $\text{ a^{2}+\dfrac{1}{a^{2}}=x }$, find the value of $a-\dfrac{1}{a}$.

$\huge\text{\underline{\underline{Solution}}}$

We know that -

$\text{ \cdots\longrightarrow\left(a-\dfrac{1}{a}\right)^{2}=a^{2}-2+\dfrac{1}{a^{2}} }$

By substitution, -

$\cdots\longrightarrow\left(a-\dfrac{1}{a}\right)^{2}=x-2$

$\cdots\longrightarrow a-\dfrac{1}{a}=\pm\sqrt{x-2}$

$\huge\text{\underline{\underline{Final answer}}}$

Hence, the value of $a-\dfrac{1}{a}$ is -

$\text{ \cdots\longrightarrow\boxed{a-\dfrac{1}{a}=\pm\sqrt{x-2}} }$

$\huge\text{\underline{\underline{Related question}}}$

$\text{ \boxed{\text{Question}} }$

If $\text{ \dfrac{x}{x^{2}+x+1}=a }$, find the value of $\dfrac{x^{2}}{x^{4}+x^{2}+1}$ in terms of $a$. $(ax\neq0)$

$\text{ \boxed{\bold{Step\ 1.}} }$

$\bold{Let\ -}$

$\text{ \cdots\longrightarrow\begin{cases} & \dfrac{x}{x^{2}+x+1}=a \\\\ & \dfrac{x^{2}}{x^{4}+x^{2}+1}=b \end{cases} }$

$\bold{Taking\ reciprocal,}$

$\text{ \cdots\longrightarrow\dfrac{x^{2}+x+1}{x}=\dfrac{1}{a} }$

$\bold{After\ division,}$

$\cdots\longrightarrow x+1+\dfrac{1}{x}=\dfrac{1}{a}$

$\bold{Manipulating,}$

$\cdots\longrightarrow x+\dfrac{1}{x}=\dfrac{1}{a}-1$

$\bold{Perfect\ square\ identity}$

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

$\text{ \cdots\longrightarrow x^{2}+2+\dfrac{1}{x^{2}}=\dfrac{1-2a+a^{2}}{a^{2}} }$

$\text{ \cdots\longrightarrow x^{2}+1+\dfrac{1}{x^{2}}=\dfrac{1-2a}{a^{2}} }$

$\text{ \boxed{\bold{Step\ 2.}} }$

$\bold{Then,\ -}$

$\text{ \cdots\longrightarrow\dfrac{x^{2}}{x^{4}+x^{2}+1}=b }$

$\bold{Taking\ reciprocal,}$

$\text{ \cdots\longrightarrow\dfrac{x^{4}+x^{2}+1}{x^{2}}=\dfrac{1}{b} }$

$\text{ \cdots\longrightarrow x^{2}+1+\dfrac{1}{x^{2}}=\dfrac{1}{b} }$

$\text{ \boxed{\bold{Step\ 3.}} }$

$\bold{Hence,\ -}$

$\text{ \cdots\longrightarrow\dfrac{1}{b}=\dfrac{1-2a}{a^{2}} }$

$\bold{Taking\ reciprocal,}$

$\text{ \cdots\longrightarrow b=\dfrac{a^{2}}{1-2a} }$

$\text{ \boxed{\bold{Final\ answer}} }$

$\text{ \cdots\longrightarrow\boxed{\dfrac{x^{2}}{x^{4}+x^{2}+1}=\dfrac{a^{2}}{1-2a}} }$

Answer 2

please check the attached file