Question

X^8-727x^4+1=0 find x-1/x

Answer 1

Answer:

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Answer 2

$\therefore (x-\frac{1}{x} )= 5$

Step-by-step explanation:

Given that,

$x^8-727x^4+1=0$

$\Rightarrow x^8 +1 = 727x^4$

The both sides are dividing by x⁴

$\Rightarrow \frac{x^8+1}{x^4} =\frac{727x^4}{x^4}$

$\Rightarrow x^4+\frac{1}{x^4} = 727$

[We know that

$(a+b)^2=a^2+2ab+b^2$

$\Rightarrow a^2+b^2 =(a+b)^2-2ab$]

Applying this in the above equation

$\Therefore (x^2+\frac{1}{x^2} )^2-2.x^2.\frac{1}{x^2} =727$

$\Rightarrow (x^2+\frac{1}{x^2} )^2-2=727$

$\Rightarrow (x^2+\frac{1}{x^2} )^2=727+2$

$\Rightarrow (x^2+\frac{1}{x^2} )^2=729$

Square root both sides

$\Rightarrow x^2+\frac{1}{x^2} =27$

[Again we know that

$(a-b)^2=a^2-2ab+b^2$

$\Rightarrow a^2+b^2 =(a-b)^2+2ab]$

Applying this in the above equation

$(x-\frac{1}{x} )^2+2.x.\frac{1}{x} =27$

$\Rightarrow (x-\frac{1}{x} )^2= 27-2$

$\Rightarrow (x-\frac{1}{x} )^2= 25$

Square root both sides

$\Rightarrow (x-\frac{1}{x} )= 5$

$\therefore (x-\frac{1}{x} )= 5$