Question

plsss answer this !! and I'll mark I the brainest and plss no spam ಥ‿ಥ , it's an expansion sum ! ​

Answer 1

$\Huge\text{(I) 5}$

$\Huge\text{(II) \pm\sqrt{29} }$

$\Huge\text{(III) \pm5\sqrt{29} }$

$\Huge\text{(IV) 27 }$

$\huge\text{\underline{\underline{Topic}}}$

$\Large\text{[Polynomials]}$

$\large\text{ \rightarrow Identity}$

What is an identity?

An identity is an equation that is always true, no matter what value we substitute into.

Here, we will be using the perfect square identity.

$\bold{Perfect\ square\ identity}$

$\text{ \cdots\longrightarrow\boxed{(x+y)^{2}=x^{2}+2xy+y^{2}} }$

$\huge\text{\underline{\underline{Explanation 27. (I)}}}$

$\bold{Given\ equation}$

$\cdots\longrightarrow a=\dfrac{1}{a-5}$

Notice that the inverse of both equations is equal.

$\cdots\longrightarrow a-5=\dfrac{1}{a}$

$\text{ \cdots\longrightarrow\boxed{a-\dfrac{1}{a}=5} }$

$\huge\text{\underline{\underline{Explanation 27. (II)}}}$

$\bold{Given\ equation}$

$\text{ \cdots\longrightarrow\boxed{a-\dfrac{1}{a}=5} }$

By squaring both sides, -

$\text{ \cdots\longrightarrow a^{2}-2+\dfrac{1}{a^{2}}=25 }$

$\text{ \cdots\longrightarrow a^{2}+2+\dfrac{1}{a^{2}}=29 }$

$\cdots\longrightarrow (a+\dfrac{1}{a})^{2}=29$

$\text{ \cdots\longrightarrow\boxed{a+\dfrac{1}{a}=\pm\sqrt{29}} }$

$\huge\text{\underline{\underline{Explanation 27. (III)}}}$

$\bold{Given\ equation}$

$\text{ \cdots\longrightarrow\boxed{\begin{cases} & a+\dfrac{1}{a}=\pm\sqrt{29} \\\\ & a-\dfrac{1}{a}= 5\end{cases}} }$

By polynomial identity -

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

$\text{ \cdots\longrightarrow\boxed{a^{2}-\dfrac{1}{a^{2}}=\pm5\sqrt{29}} }$

$\huge\text{\underline{\underline{Explanation 27. (IV)}}}$

$\bold{Given\ equation}$

$\text{ \cdots\longrightarrow\boxed{a-\dfrac{1}{a}=5} }$

By squaring both sides -

$\text{ \cdots\longrightarrow a^{2}-2+\dfrac{1}{a^{2}}=25 }$

$\text{ \cdots\longrightarrow\boxed{a^{2}+\dfrac{1}{a^{2}}=27} }$

In these types of problems, knowing the appropriate identity to solve the problem is important. Let's keep learning, I hope it helps!