Question

$\sf\ If\ \ abx^2= (a-b)^2(x+1)\\ \\ \sf\ then\ find\ the \ value\ of\ \\ \\ \sf\ x+\dfrac{4}{x^2}+\dfrac{4}{x}$​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Algebraic Identities has been used . We see we are given a equation from that we have to find other . So firstly we can simplify the value of first equation and then apply in second .

Let's do it !!

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★ Correct Question :-

If abx² = (a - b)² (x + 1) then find,

$\\\;\sf{\pink{\mapsto\;\;1\;+\;\dfrac{4}{x^{2}}\;+\;\dfrac{4}{x}}}$

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★ Solution :-

Given,

$\\\;\sf{\orange{:\rightarrow\;\;abx^{2}\;=\;\bf{(a\;-\;b)^{2}\:(x\;+\;1)}}}$

• Now simplifying this value, we get,

$\\\;\sf{:\Longrightarrow\;\;abx^{2}\;=\;\bf{(a\;-\;b)^{2}\:(x\;+\;1)}}$

• Transposing similar terms to other side, we get,

$\\\;\sf{:\Longrightarrow\;\;\dfrac{ab}{\blue{(a\;-\;b)^{2}}}\;=\;\bf{\dfrac{(x\;+\;1)}{\blue{x^{2}}}}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{ab}{(a\;-\;b)^{2}}\;=\;\bf{\bigg(\dfrac{x}{x^{2}}\bigg)\;+\;\bigg(\dfrac{1}{x^{2}}\bigg)}}$

• Cancelling x, we get

$\\\;\sf{:\Longrightarrow\;\;\green{\dfrac{ab}{(a\;-\;b)^{2}}}\;=\;\bf{{\bigg(\dfrac{1}{x}\;+\;\dfrac{1}{x^{2}}\bigg)}}}$

Let this be equation i) .

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~ To find the required value ::

Its given that,

$\\\;\sf{\rightarrow\;\;1\;+\;\dfrac{4}{x}\;+\;\dfrac{4}{x^{2}}}$

$\\\;\sf{\Longrightarrow\;\;1\;+\;\bf{\bigg(\dfrac{4}{x}\;+\;\dfrac{4}{x^{2}}\bigg)}}$

• Here we see that 4 is a common term in both fractions. So taking that in common we get,

$\\\;\sf{\Longrightarrow\;\;1\;+\;\bf{\red{4}\:\bigg(\dfrac{1}{x}\;+\;\dfrac{1}{x^{2}}\bigg)}}$

• We see that term here and in equation i) is common. So applying that value, we get,

$\\\;\sf{\Longrightarrow\;\;1\;+\;\bf{4\:\bigg(\dfrac{ab}{(a\;-\;b)^{2}}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;1\;+\;\bf{\bigg(4\:\times\:\dfrac{ab}{(a\;-\;b)^{2}}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;1\;+\;\bf{\bigg(4\:\times\:\dfrac{ab}{(a\;-\;b)^{2}}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;1\;+\;\bf{\bigg(\dfrac{4ab}{(a\;-\;b)^{2}}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;1\;+\;\bf{\dfrac{4ab}{(a\;-\;b)^{2}}}}$

$\\\;\bf{\Longrightarrow\;\;\dfrac{1(a\;-\;b)^{2}\;+\;4ab}{(a\;-\;b)^{2}}}$

$\\\;\bf{\Longrightarrow\;\;\dfrac{(a^{2}\;+\;b^{2}\;-\;2ab)\;+\;4ab}{(a\;-\;b)^{2}}}$

$\\\;\bf{\Longrightarrow\;\;\dfrac{a^{2}\;+\;b^{2}\;-\;2ab\;+\;4ab}{(a\;-\;b)^{2}}}$

$\\\;\bf{\Longrightarrow\;\;\dfrac{a^{2}\;+\;b^{2}\;+\;2ab}{(a\;-\;b)^{2}}}$

This forms the identity of (a + b)² .

$\;\bf{\Longrightarrow\;\;\dfrac{(a\;+\;b)^{2}}{(a\;-\;b)^{2}}}$

$\\\;\bf{\purple{\Longrightarrow\;\;\bigg(\dfrac{(a\;+\;b)}{(a\;-\;b)}\bigg)^{2}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;required\;\;answer\;=\;\bf{\purple{\bigg(\dfrac{(a\;+\;b)}{(a\;-\;b)}\bigg)^{2}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;(a\;-\;b)^{2}\;=\;a^{2}\;+\;b^{2}\;-\;2ab}$

$\\\;\sf{\leadsto\;\;a^{2}\;-\;b^{2}\;=\;(a\;+\;b)(a\;-\;b)}$

$\\\;\sf{\leadsto\;\;(x\;+\;a)(x\;+\;b)\;=\;x^{2}\;+\;(a\;+\;b)x\;+\;ab}$

$\\\;\sf{\leadsto\;\;(a\;+\;b)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;2ab}$

$\\\;\sf{\leadsto\;\;(a\;+\;b\;+\;c)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;c^{2}\;+\;2ab\;+\;2bc\;+\;2ac}$