Question

if the area of a square is
${x}^{2} + \frac{1}{ {x}^{2} } + 3 - 2x - \frac{2}{x}$
find it's one side​

Answer 1

Answer:

$(x + \frac{1}{x} - 1)(x + \frac{1}{x} - 1)$

Step-by-step explanation:

Area of the square given =

x² + 1 +1+1/x²-x-x-1/x-1/x

As we know the area of square = side²

so, side =√x² + 1 +1+1/x²-x-x-1/x-1/x

Solution :

• ${x}^{2} + \frac{1}{ {x}^{2} } + 3 - 2x - \frac{2}{x}$

${x}^{2} + 1 + 1 + 1 + \frac{1}{ {x}^{2} } -x - x - \frac{1}{x} - \frac{1}{x}$

${x}^{2} + 1 - x + 1 + \frac{1}{ {x}^{2} } - \frac{1}{x} + 1 - \frac{1}{x} - \frac{1}{x}$

$x(x + \frac{1}{x} - 1) + \frac{1}{x} (x + \frac{1}{x} - 1) - 1(x + \frac{1}{x} - 1)$

$= > (x + \frac{1}{x} - 1)(x + \frac{1}{x} - 1)$