Observe that
[tex]\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = \left(\sqrt x\right)^2 - 2\sqrt x\dfrac1{\sqrt x} + \left(\dfrac1{\sqrt x}\right)^2 = x - 2 + \dfrac1x[/tex]
Now,
[tex]x = 9 - 4\sqrt5 \implies \dfrac1x = \dfrac1{9-4\sqrt5} = \dfrac{9 + 4\sqrt5}{9^2 - \left(4\sqrt5\right)^2} = 9 + 4\sqrt5[/tex]
so that
[tex]\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = (9 - 4\sqrt5) - 2 + (9 + 4\sqrt5) = 16[/tex]
[tex]\implies \sqrt x - \dfrac1{\sqrt x} = \pm\sqrt{16} = \pm 4[/tex]
To decide which is the correct value, we need to examine the sign of [tex]\sqrt x - \frac1{\sqrt x}[/tex]. It evaluates to 0 if
[tex]\sqrt x = \dfrac1{\sqrt x} \implies x = 1[/tex]
We have
[tex]9 - 4\sqrt5 = \sqrt{81} - \sqrt{16\cdot5} = \sqrt{81} - \sqrt{80} > 0[/tex]
Also,
[tex]\sqrt{81} - \sqrt{64} = 9 - 8 = 1[/tex]
and [tex]\sqrt x[/tex] increases as [tex]x[/tex] increases, which means
[tex]0 < 9 - 4\sqrt5 < 1[/tex]
Therefore for all [tex]0 < x < 1[/tex],
[tex]\sqrt x - \dfrac1{\sqrt x} < 0[/tex]
For example, when [tex]x=\frac14[/tex], we get
[tex]\sqrt{\dfrac14} - \dfrac1{\sqrt{\frac14}} = \dfrac1{\sqrt4} - \sqrt4 = \dfrac12 - 2 = -\dfrac32 < 0[/tex]
Then the target expression has a negative sign at the given value of [tex]x[/tex] :
[tex]x = 9-4\sqrt5 \implies \sqrt x - \dfrac1{\sqrt x} = \boxed{-4}[/tex]
Alternatively, we can try simplifying [tex]\sqrt x[/tex] by denesting the radical. Let [tex]a,b,c[/tex] be non-zero integers ([tex]c>0[/tex]) such that
[tex]\sqrt{9 - 4\sqrt5} = a + b\sqrt c[/tex]
Note that the left side must be positive.
Taking squares on both sides gives
[tex]9 - 4\sqrt5 = a^2 + 2ab\sqrt c + b^2c[/tex]
Let [tex]c=5[/tex] and [tex]ab=-2[/tex]. Then
[tex]a^2+5b^2=9 \implies a^2 + 5\left(-\dfrac2a\right)^2 = 9 \\\\ \implies a^2 + \dfrac{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[/tex]
[tex]a^2 = 4 \implies 5b^2 = 5 \implies b^2 = 1[/tex]
[tex]a^2 = 5 \implies 5b^2 = 4 \implies b^2 = \dfrac45[/tex]
Only the first case leads to integer coefficients. Since [tex]ab=-2[/tex], one of [tex]a[/tex] or [tex]b[/tex] must be negative. We have
[tex]a^2 = 4 \implies a = 2 \text{ or } a = -2[/tex]
Now if [tex]a=2[/tex], then [tex]b=-1[/tex], and
[tex]\sqrt{9 - 4\sqrt5} = 2 - \sqrt5[/tex]
However, [tex]\sqrt5 > \sqrt4 = 2[/tex], so [tex]2-\sqrt5[/tex] is negative, so we don't want this.
Instead, if [tex]a=-2[/tex], then [tex]b=1[/tex], and thus
[tex]\sqrt{9 - 4\sqrt5} = -2 + \sqrt5[/tex]
Then our target expression evaluates to
[tex]\sqrt x - \dfrac1{\sqrt x} = -2 + \sqrt5 - \dfrac1{-2 + \sqrt5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - \dfrac{-2 - \sqrt5}{(-2)^2 - \left(\sqrt5\right)^2} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 + \dfrac{2 + \sqrt5}{4 - 5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - (2 + \sqrt5) = \boxed{-4}[/tex]
