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Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

This solution is from Arun Iyer, SIA High School and Junior College.

The given equation is $x^{3/2} - 8x^{-3/2} = 7$.

Multiply throughout by $x^{3/2}$and rearranging we get

$(x^{3/2})^2 - 7x^{3/2} - 8 = 0$

This can be factorised as:

$(x^{3/2} - 8)(x^{3/2} + 1) = 0$

Case 1:

Consider,

$x^{3/2} + 1 = 0$ or $x^{3/2} = -1$

Squaring both sides, $x^3 = 1$ so the three cube roots of unity are solutions.

Now

$ 1 = e^{2k\pi i} $

(for $k=0,1,2,3$...) therefore $ x = e^{{2k\pi i} /3} $ .

Putting $k=0$, $1$, $2$ we get:

$x = e^0$ (which is $1$) or

$ x = exp({\pm({2\pi i} /3)}) $

.

Case 2:

Consider,

$x^{3/2}- 8 = 0$ or $x^{3/2} = 8$

Squaring both sides, $x^3 = 64$ so the three cube roots of $64$ are solutions.

Now

$ x^3 = 64 e^{2k\pi i} $

(for $k=0,1,2,3$...) therefore

$ x = 64^{1/3} e^{{2k\pi i}/3} $ .

Putting $k=0,1,2$ we get: $x = 4$ or

$ x = 4exp({\pm{2\pi i} /3}) $

.

So the six roots are:

  • $x_1 = 1$
  • $x_2 = \exp({2\pi i} /3) = \cos (2\pi /3) + i\sin(2\pi /3) = -1/2 + i {\sqrt3}/2 $
  • $x_3 = \exp({-2\pi i} /3) = \cos (-2\pi /3) + i\sin(-2\pi /3) = -1/2 - i {\sqrt3}/2 $
  • $x_4 = 4$
  • $x_5 = 4\exp({2\pi i} /3) = 4\cos (2\pi /3) + 4i\sin(2\pi /3) = -2 + 2i {\sqrt3} $
  • $x_6 = 4\exp({-2\pi i} /3) = 4\cos (-2\pi /3) + 4i\sin(-2\pi /3) = -2 - 2i {\sqrt3} $

Substituting $x=1$ into the given equation we need to recognise that $1^{3/2}$ has two values $+1$ and $-1$ so that whereas $x^{3/2} = 1$ does not satisfy the equation the other value $x^{3/2} = -1$ does satisfy it and hence $x=1$ is a solution of the equation.