Question

Solving Equation
(x-1/x)^2-6(x+1/x) +12 =0
we get roots as ?

Please Give Full Solution.

(b) 1 is Correct Answer​

Answer 1

Step-by-step explanation:

this is the solution of the above equation .

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Answer 2

The values of x are 1 and $\frac{4 \pm \sqrt{12} }{2}$

Given:

$(x-\frac{1}{x})^{2} -6(x+\frac{1}{x})+12 = 0$

To Find:

The roots of the given equation

Solution:

Let $P=(x+\frac{1}{x})$

Squaring on both sides

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

Given that,

$(x-\frac{1}{x})^{2} -6(x+\frac{1}{x})+12 = 0$

$x^{2} -2+\frac{1}{x^{2} } -6P+12 = 0\\\\x^{2} + \frac{1}{x^{2} } -2-6P+12=0\\\\P^{2} -2-2-6P+12 = 0-------------(Using 1)\\\\P^{2} -6P +8 =0\\\\P^{2}-4P-2P +8 =0\\ \\P(P-4)-2(P-4) = 0\\\\(P-2)(P-4)=0\\\\(P-2) = 0 \ or \ (P-4)=0\\\\P=2 \ or\ P=4$

   Substituting the value of P.

$2=(x+\frac{1}{x}) \ or \ 4=(x+\frac{1}{x})\\\\2= \frac{x^{2} +1}{x} \ or \ 4 = \frac{x^{2}+1 }{4}\\\\2x=x^{2} +1 \ or \ 4x=x^{2} +1\\\\x^{2} -2x+1 \ or \ x^{2} -4x+1$

Case i)

$x^{2} -2x+1=0\\\\ \Here a=1,b=-2 \ and c=1\\\\x = \frac{-(-2) \pm \sqrt{-(2)^{2}-4 \times 1 \times 1 } }{2} \\\\x= \frac{2 \pm \sqrt{0} }{2} \\\\x= \frac{2}{2} \\\\x=1$

Case ii)

$x^{2} -4x+1 =0\\\\\ Here a=1,b=-4 \ and \ c=1\\\\x=\frac{-(-4)\pm\sqrt{(-4)^{2} -4 \times 1 \times 1} }{2} \\\\x= \frac{4 \pm \sqrt{16-4} }{2} \\\\x =\frac{4 \pm \sqrt{12} }{2}$

Therefore, the values of x are 1 and $\frac{4 \pm \sqrt{12} }{2}$

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