If (x + 1/x)^2 =3 , then (x^{90} + 1/x^{90}) =?

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Apr 21, 2026, 1:02:34 AM (8 days ago) Apr 21

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Mathematics for School & Entrance Level • 1.5K followers

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Ragu Rajagopalan, Passionate Maths solver ;Reviving knowledge after 3 decades

[math]\\[/math]

[math]\left(x + \dfrac{1}{x} \right)^2 = 3 \implies x^2 + \dfrac{1}{x^2} + 2 = 3 [/math]

[math]\implies x^2 + \dfrac{1}{x^2} = 1 \implies x^4 - x^2 + 1 = 0 [/math]

[math]\implies (x^2 + 1) \cdot (x^4 - x^2 + 1) = 0 \implies (x^6 + 1) = 0 [/math]

[math]\implies x^6 = -1 \\[/math]

[math]\implies x^{90} = (x^6)^{15} = (-1)^{15} = -1 [/math]

[math]\therefore x^{90} + \dfrac{1}{x^{90}} = -1- 1 = \boxed{\boxed{-\: 2 \:}}[/math]

[math]18–10,\ 54+10,\ 90–10,\ 126+10,\ 162–10,\ldots,\ 18(2n-1)+10(-1)^n,\ldots[/math]

[math]\therefore\,\boxed{T_{100}=18(200-1)+10(-1)^{100}=3582+10=3592}\,\blacksquare[/math]

Pradeep Hebbar, Many years of Structural Engineering & Math enthusiasm

Given

[math]\dfrac{3x+5}{x+2} = 2+\dfrac{x+2}{x-3}[/math]

We write,

[math]\dfrac{3x+5}{x+2}-\dfrac{x+2}{x-3} = 2[/math]

[math]\dfrac{(3x+5)(x-3)-(x+2)(x+2)}{(x+2)(x-3)} = 2[/math]

[math]\dfrac{2x^2-8x-19}{x^2-x-6}=2[/math]

[math]2x^2-8x-19= 2(x^2-x-6)[/math]

[math]-6x-7 =0 [/math]

[math]x= -\frac{7}{6}[/math]

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