Question

If
`x = 9 - 4√(5)`, find the value of
`√(x) - (1)/(√(x) )`.

Answer 1

Observe that

`\left(\sqrt x - \frac1{\sqrt x}\right)^2 = \left(\sqrt x\right)^2 - 2\sqrt x\frac1{\sqrt x} + \left(\frac1{\sqrt x}\right)^2 = x - 2 + \frac1x`

Now,

`x = 9 - 4\sqrt5 \implies \frac1x = \frac1{9-4\sqrt5} = (9 + 4\sqrt5)/(9^2 - \left(4\sqrt5\right)^2) = 9 + 4\sqrt5`

so that

`\left(\sqrt x - \frac1{\sqrt x}\right)^2 = (9 - 4\sqrt5) - 2 + (9 + 4\sqrt5) = 16`

`\implies \sqrt x - \frac1{\sqrt x} = \pm√(16) = \pm 4`

To decide which is the correct value, we need to examine the sign of
`\sqrt x - \frac1{\sqrt x}`. It evaluates to 0 if

`\sqrt x = \frac1{\sqrt x} \implies x = 1`

We have

`9 - 4\sqrt5 = √(81) - √(16\cdot5) = √(81) - √(80) > 0`

Also,

`√(81) - √(64) = 9 - 8 = 1`

and
`\sqrt x` increases as
`x` increases, which means

`0 < 9 - 4\sqrt5 < 1`

Therefore for all
`0 < x < 1`,

`\sqrt x - \frac1{\sqrt x} < 0`

For example, when
`x=\frac14`, we get

`√(\frac14) - \frac1{√(\frac14)} = \frac1{\sqrt4} - \sqrt4 = \frac12 - 2 = -\frac32 < 0`

Then the target expression has a negative sign at the given value of
`x` :

`x = 9-4\sqrt5 \implies \sqrt x - \frac1{\sqrt x} = \boxed{-4}`

Alternatively, we can try simplifying
`\sqrt x` by denesting the radical. Let
`a,b,c` be non-zero integers (
`c>0`) such that

`√(9 - 4\sqrt5) = a + b\sqrt c`

Note that the left side must be positive.

Taking squares on both sides gives

`9 - 4\sqrt5 = a^2 + 2ab\sqrt c + b^2c`

Let
`c=5` and
`ab=-2`. Then

`a^2+5b^2=9 \implies a^2 + 5\left(-\frac2a\right)^2 = 9 \\\\ \implies a^2 + (20)/(a^2) = 9 \\\\ \implies a^4 + 20 = 9a^2 \\\\ \implies a^4 - 9a^2 + 20 = 0 \\\\ \implies (a^2 - 4) (a^2 - 5) = 0 \\\\ \implies a^2 = 4 \text{ or } a^2 = 5`

`a^2 = 4 \implies 5b^2 = 5 \implies b^2 = 1`

`a^2 = 5 \implies 5b^2 = 4 \implies b^2 = \frac45`

Only the first case leads to integer coefficients. Since
`ab=-2`, one of
`a` or
`b` must be negative. We have

`a^2 = 4 \implies a = 2 \text{ or } a = -2`

Now if
`a=2`, then
`b=-1`, and

`√(9 - 4\sqrt5) = 2 - \sqrt5`

However,
`\sqrt5 > \sqrt4 = 2`, so
`2-\sqrt5` is negative, so we don't want this.

Instead, if
`a=-2`, then
`b=1`, and thus

`√(9 - 4\sqrt5) = -2 + \sqrt5`

Then our target expression evaluates to

`\sqrt x - \frac1{\sqrt x} = -2 + \sqrt5 - \frac1{-2 + \sqrt5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - (-2 - \sqrt5)/((-2)^2 - \left(\sqrt5\right)^2) \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 + (2 + \sqrt5)/(4 - 5) \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - (2 + \sqrt5) = \boxed{-4}`