Question

Solve the equation : $\bf \: x \times \dfrac{1}{x} = 2 \times \dfrac{1}{20} \: \: x \neq0$​

Answer 1

Appropriate Question :-

Solve the equation :-

$\rm \: x\dfrac{1}{x} = 2 \frac{1}{20}$

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

Given equation is

$\rm \: x\dfrac{1}{x} = 2 \frac{1}{20}$

can be rewritten as

$\rm \: \frac{ {x}^{2} + 1 }{x} = \frac{40 + 1}{20}$

$\rm \: \frac{ {x}^{2} + 1 }{x} = \frac{41}{20}$

$\rm \: 20( {x}^{2} + 1) = 41x$

$\rm \: 20 {x}^{2} + 20= 41x$

$\rm \: 20 {x}^{2} - 41x + 20= 0$

$\rm \: 20 {x}^{2} - 25x - 16x + 20= 0$

$\rm \: 5x(4x - 5) - 4(4x - 5) = 0$

$\rm \: (4x - 5)(5x - 4) = 0$

$\rm\implies \:x = \frac{4}{5} \: \: \: or \: \: \: x = \frac{5}{4}$

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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

$\underline{ \bf{GIVEN } }:$

$\rm \qquad \mapsto \: \: x \frac{1}{x} = 2 \frac{1}{20}$

$\underline{ \mathbf{SOLUTION  }} :$

$\implies \rm \frac{ {x}^{2} + 1}{x} = \frac{20 \times 2 + 1}{20}$

$\implies \rm20( {x}^{2} + 1) = 41(x)$

$\implies \rm20 {x}^{2} + 20 = 41x$

$\implies \rm20 {x}^{2} - 41x + 20 = 0$

$\implies \rm20 {x}^{2} - 25x - 16x + 20 = 0$

$\implies \rm(4x - 5)(5x - 4)$

$\implies \rm \: x = \frac{5}{4} , \frac{4}{5}$