The number of integral solutions of 2(x2+1x2)−7(x+1x)+9=0 when x≠0 is

Given 2(x2+1x2)−7(x+1x)+9=0Let x+1x=9⇒ x2+1x2=a2−2⇒ 2(a2−2)−7a+9=0⇒ 2a2−4−7a+9=0⇒ 2a2−7a+5=0∴ 2a2−5a−2a+5=0⇒ a(2a−5)−1(2a−5)=0⇒ (2a−5)(a−1)=0⇒ a=52, a=1case(i): If a=1 then x+1x=1⇒ x2+1=x⇒ x2−x+1=0⇒ x= 1±1−42⇒ x= 1±i32If a=52 then x+1x=52⇒ x2+1x=52⇒ 2x2+2=5x⇒ 2x2−5x+2=0⇒ 2x2−4x+2=0⇒ 2x(x−2)−1(x−2)=0⇒ (x−2)(2x−1)=0x=2, x=12number of integral solutions are 1

The number of integral solutions of 2(x2+1x2)−7(x+1x)+9=0 when x≠0 is

Given 2(x2+1x2)−7(x+1x)+9=0Let x+1x=9⇒ x2+1x2=a2−2⇒ 2(a2−2)−7a+9=0⇒ 2a2−4−7a+9=0⇒ 2a2−7a+5=0∴ 2a2−5a−2a+5=0⇒ a(2a−5)−1(2a−5)=0⇒ (2a−5)(a−1)=0⇒ a=52, a=1case(i): If a=1 then x+1x=1⇒ x2+1=x⇒ x2−x+1=0⇒ x= 1±1−42⇒ x= 1±i32If a=52 then x+1x=52⇒ x2+1x=52⇒ 2x2+2=5x⇒ 2x2−5x+2=0⇒ 2x2−4x+2=0⇒ 2x(x−2)−1(x−2)=0⇒ (x−2)(2x−1)=0x=2, x=12number of integral solutions are 1

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$T h e n u m b e r o f int e g r a l s o l u t i o n s o f 2 (x^{2} + \frac{1}{x^{2}}) - 7 (x + \frac{1}{x}) + 9 = 0 w h e n x \neq 0 i s$

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Detailed Solution

$\begin{array}{l} G i v e n 2 (x^{2} + \frac{1}{x^{2}}) - 7 (x + \frac{1}{x}) + 9 = 0 \\ L e t x + \frac{1}{x} = 9 \Rightarrow x^{2} + \frac{1}{x^{2}} = a^{2} - 2 \\ \Rightarrow 2 (a^{2} - 2) - 7 a + 9 = 0 \\ \Rightarrow 2 a^{2} - 4 - 7 a + 9 = 0 \\ \Rightarrow 2 a^{2} - 7 a + 5 = 0 \\ ∴ 2 a^{2} - 5 a - 2 a + 5 = 0 \\ \Rightarrow a (2 a - 5) - 1 (2 a - 5) = 0 \\ \Rightarrow (2 a - 5) (a - 1) = 0 \\ \Rightarrow a = \frac{5}{2}, a = 1 \\ c a s e (i) : I f a = 1 t h e n x + \frac{1}{x} = 1 \\ \Rightarrow x^{2} + 1 = x \\ \Rightarrow x^{2} - x + 1 = 0 \\ \Rightarrow x = \frac{1 \pm \sqrt{1 - 4}}{2} \\ \Rightarrow x = \frac{1 \pm i \sqrt{3}}{2} \\ I f a = \frac{5}{2} t h e n x + \frac{1}{x} = \frac{5}{2} \\ \Rightarrow \frac{x^{2} + 1}{x} = \frac{5}{2} \\ \Rightarrow 2 x^{2} + 2 = 5 x \\ \Rightarrow 2 x^{2} - 5 x + 2 = 0 \\ \Rightarrow 2 x^{2} - 4 x + 2 = 0 \\ \Rightarrow 2 x (x - 2) - 1 (x - 2) = 0 \\ \Rightarrow (x - 2) (2 x - 1) = 0 \\ x = 2, x = \frac{1}{2} \\ n u m b e r o f int e g r a l s o l u t i o n s a r e 1 \end{array}$