Question

Solve using method of factorization
$x + \frac{1}{x} = 25 \frac{1}{25}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm \: x + \dfrac{1}{x} = 25\dfrac{1}{25}$

can be rewritten as

$\rm \: \dfrac{ {x}^{2} + 1}{x} = 25 + \dfrac{1}{25}$

$\rm \: {x}^{2} + 1 = 25x + \dfrac{1}{25}x$

can be rewritten as

$\rm \: {x}^{2} + 1 - 25x - \dfrac{1}{25}x = 0$

can be re-arranged as

$\rm \: {x}^{2} - 25x - \dfrac{1}{25}x + 1 = 0$

can be further rewritten as

$\rm \: {x}^{2} - 25x - \dfrac{1}{25}x + 25 \times \dfrac{1}{25} = 0$

$\rm \: x(x - 25) - \dfrac{1}{25}(x - 25) = 0$

$\rm \: (x - 25)\bigg(x - \dfrac{1}{25} \bigg) = 0$

$\rm\implies \:x = 25 \: \: or \: \: x = \dfrac{1}{25}$

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Short Cut trick :-

$\boxed{\sf{  \:\rm \: If \: x + \frac{1}{x} = y\frac{1}{y} \: \: \: \: then \: x = y \: or \: x = \frac{1}{y} \: \: }}$

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Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

Step-by-step explanation:

This is a simple, short and memorable way.

$\pmb{\[ \begin{array}{l} \color{darkviolet} \tt x+\dfrac{1}{x}=25 \dfrac{1}{25} \\ \\ \tt \Rightarrow \red x+ \orange{\dfrac{1}{x}}= \red{25}+ \orange{\dfrac{1}{25}} \\ \\ \color{darkviolet} \tt \left(\because a \dfrac{b}{c}=a+\dfrac{b}{c}\right) \end{array} \]}$

on comparing, we get:-

$\color{magenta}\[ \begin{array}{l} \tt \implies x=25 \: \: \: \: \: and \\ \implies \tt x= \dfrac{1}{25} \end{array} \]$