Question

Solve : (x + 1/x)^2 - 3/2 ( x - 1/x) = 4

Answer 1

Answer:

Hey... here is your answer in the image.

Hope it helps! ! !

Answer 2

The correct answers are -1, 1, 2 and $\frac{-1}{2}$.

Given: The equation = $(x+\frac{1}{x} )^{2} -\frac{3}{2} (x-\frac{1}{x} )=4$.

To Find: Evaluate the equation.

Solution:

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

Add and subtract $2*x*\frac{1}{x}$.

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

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

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

Put this in the original equation.

$(x+\frac{1}{x} )^{2} -\frac{3}{2} (x-\frac{1}{x} )=4$

$(x-\frac{1}{x} )^{2} +4-\frac{3}{2} (x-\frac{1}{x} )=4$

Let $x-\frac{1}{x} =p$.

$p^{2} -\frac{3}{2}p=0$

$p(p-\frac{3}{2} )=0$

p = 0 and p = $\frac{3}{2}$.

Firstly take p = 0.

$x-\frac{1}{x}=0$

$x^{2} -1=0$

(x-1)(x+1) = 0

x = +1 and -1.

Now take $p=\frac{3}{2}$.

$x-\frac{1}{x} =\frac{3}{2}$

$\frac{x^{2}-1 }{x} =\frac{3}{2}$

$2x^{2} -2=3x$

$2x^{2} -3x-2=0$

$2x^{2} -4x+x-2=0\\ 2x(x-2)+(x-2)$

(2x+1)(x-2)

x = 2, $\frac{-1}{2}$

Hence, the values of x are -1, 1, 2 and $\frac{-1}{2}$.

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