Question

Find the value(s) of x using factorization method

${x}+ \frac{1}{x} = 25 \frac{1}{25}$
​

Answer 1

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

Given expression is

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

can be rewritten as

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

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

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

can be further rewritten as

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

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

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

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

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SHORT CUT TRICK

☆ If the equation is of the form

$\rm :\longmapsto\:x + \dfrac{1}{x} = a\dfrac{1}{a} \: then \: \\ \red{\rm\implies \:x = a \: \: or \: x = \dfrac{1}{a} }$

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More to Know

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

Answer:

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

Given expression is

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

can be rewritten as

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

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

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

can be further rewritten as

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

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

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

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

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

SHORT CUT TRICK

☆ If the equation is of the form

$\rm :\longmapsto\:x + \dfrac{1}{x} = a\dfrac{1}{a} \: then \: \\ \red{\rm\implies \:x = a \: \: or \: x = \dfrac{1}{a} }$

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$