Can they be equal ?

Algebra Level 2

$\begin{aligned} x^2 +\dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} \end{aligned}$ The above equation is true for $x=1$ . Does the above equation have any other real solutions besides $x=1$ ?

Yes No

3 solutions

Chew-Seong Cheong
Mar 23, 2018

$\begin{aligned} x^2 + \frac 1{x^2} & = x^3 + \frac 1{x^3} & \small \color{#3D99F6} \text{Multiply both sides by }x^3 \\ x^5 + x & = x^6 + 1 & \small \color{#3D99F6} \text{Rearrange} \\ x^6-x^5-x+1 & = 0 \\ (x-1)(x^5-1) & = 0 \\ \implies x & = 1 \end{aligned}$ .

No , there is no other solution besides $x=1$ .

Actually I tried differently and concluded my ans. ..... I transposed x^3 ND 1/x^3 to LHS ...ND took lcm...ND got the eq.

x^9 -x^8 -x^4 +x^2 =0

This eq.have 9 roots out of which one is 1. So if one sol. Is 1 ..then the other 8 are complex solutions ..as they always occur in pair.....ND if not there must be 2more real sol. to maintain the pair in complex roots

Aditya Anand - 3 years, 2 months ago

No, I think you got it wrong. It should be $x^9-x^8-x^4+x^3 = 0$ and it does not means it has 9 complex roots including 1 because dividing it by $x*3$ throughout we have $x^6-x^5-x+1 = 0$ . Otherwise $x^{10}-x^9-x^5+x^4 = 0$ will have 10 roots and so on.

Chew-Seong Cheong - 3 years, 2 months ago

Wonderfull!

Heder Oliveira Dias - 3 years, 1 month ago

Munem Shahriar
Mar 22, 2018

$\begin{aligned} x^3 + \dfrac 1{x^3} & = x^2 + \dfrac 1{x^2} \\ \dfrac{x^6 + 1}{x^3} & = \dfrac{x^4 + 1}{x^2} \\ x^2(x^6 +1) & = x^3(x^4+ 1) ~~~~~~[\text{Cross multiplying}] \\ x^8 + x^2 & = x^7 + x^3 \\ x^2(x^6 + 1) & = x^2(x^5 + x)\\ x^6 +1& = x^5 + x \\ x^6 - x^5 - x + 1 & = 0 \\ x^5(x - 1) - (x - 1) & = 0 \\ (x -1) (x^5 - 1) & = 0 \implies x = 1\\ \end{aligned}$

So there is no real solution other than 1.

康鈞翁
May 31, 2019

Let t=x+1/x,then t^2-2=t^3-3t So t^3-t^2-3t+2=0 and the Ed just exactly one real solution that t=1.

Can they be equal ?

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