Question

x²+1/x² = 51 find x​

Answer 1

Answer:

let try

Step-by-step explanation:

x^4+1=51x^2

x^4-51x^2+1=0

x^4-50x^2-1x^2+1=0

Answer 2

We have,

$\qquad \: \: {x}^{2} + \frac{1}{ {x}^{2} } = 51 \\ \\ \\ \implies {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 49 \\ \\ \\ \implies {(x)}^{2} + { \left( \frac{1}{x} \right) }^{2} - 2 \times x \times \frac{1}{x} = 7 \times 7 \\ \\ \\ \implies { \left( x + \frac{1}{x} \right) }^{2} = {(7)}^{2} \\ \\ \\ \implies x + \frac{1}{x} = 7 \\ \\ \\ \implies \frac{ {x}^{2} + 1}{x} = 7 \\ \\ \\ \implies {x}^{2} + 1 = 7x \\ \\ \\ \implies {x}^{2} - 7x + 1 = 0$
This equation is of the form : $\boxed{a {x}^{2} + bx + c = 0}$

Here,

$D = \sqrt{ {b}^{2} - 4ac} \\ \\ = \sqrt{ {( - 7)}^{2} - 4(1)} \\ \\ = \sqrt{49 - 4} \\ \\ = \sqrt{45} \\ \\ = 3 \sqrt{5} \: \: > \: \: 0$
Thus, the zeroes are real and equal.

Hence,
$x = \frac{ - b \: \pm \: \sqrt{D} }{2a} \\ \\ = \frac{7 \pm 3 \sqrt{5} }{2}$
Henceforth,
$\\ x = \boxed{ \frac{7 + 3 \sqrt{5} }{2} } \quad \: or \: \quad \boxed{ \frac{7 - 3 \sqrt{5} }{2} }$