The number of integral solutions of 2(x^2 + f | vedclass

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Question

(A) Let $t = x + \frac{1}{x}$. Then $t^2 = x^2 + \frac{1}{x^2} + 2$,so $x^2 + \frac{1}{x^2} = t^2 - 2$.Substituting this into the equation: $2(t^2 - 2

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$1$

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(A) Let $t = x + \frac{1}{x}$. Then $t^2 = x^2 + \frac{1}{x^2} + 2$,so $x^2 + \frac{1}{x^2} = t^2 - 2$.Substituting this into the equation: $2(t^2 - 2) - 7t + 9 = 0$.$2t^2 - 4 - 7t + 9 = 0 \implies 2t^2 - 7t + 5 = 0$.Factoring the quadratic: $(2t - 5)(t - 1) = 0$.So,$t = 1$ or $t = \frac{5}{2}$.Case $1$: $x + \frac{1}{x} = 1 \implies x^2 - x + 1 = 0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 Case $2$: $x + \frac{1}{x} = \frac{5}{2} \implies 2x^2 - 5x + 2 = 0$.Factoring: $(2x - 1)(x - 2) = 0$,so $x = \frac{1}{2}$ or $x = 2$.Since the question asks for the number of integral solutions,we check $x = 2$ (which is an integer) and $x = \frac{1}{2}$ (which is not).Thus,there is only $1$ integral solution,which is $x = 2$.

The number of integral solutions of $2(x^2 + \frac{1}{x^2}) - 7(x + \frac{1}{x}) + 9 = 0$ for $x \neq 0$ is:

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