How many real solutions?

Trial and error shows that $x=4$ is a solution. I will attempt to prove that there are two other real roots. $$ Let $y = \sqrt{8-x} ,$ then the equation becomes $(8-y^2)^2 (y-1) = 8y$ . Expanding it gives $y^5 -y^4 - 16y^3 + 16y^2 + 56y - 64 = 0$ Since $y = \sqrt{8-x}$ , we're only interested in finding the positive roots of the quartic polynomial $f(y) := y^5 -y^4 - 16y^3 + 16y^2 + 56y - 64 .$ One of which is $y =\sqrt{8-4} = 2$ . $$ By Descartes' rule of signs , $f(y)$ have exactly 1 or 3 positive roots. Because $f(3) < 0 < f(4)$ , there lies another root of $f(y)$ in the interval $(3,4)$ , thus we've found another positive root of $f(y)$ . This means that $f(y)$ must have 3 positive roots, and so the original equation must have $\boxed3$ real solutions.

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If the author really did intend for us to solve the original equation... $$ Since $f(y)$ has a factor of $(y-2)$ , by long division, $f(y) = (y-2) (y^4 + y^3 - 14 y^2 - 12 y + 32)$ . Rational root theorem tells us that the quartic factor has no rational root. Attempting to factor the quartic polynomial as a product of two quadratic polynomials: $y^4 + y^3 - 14 y^2 - 12 y + 32 = (y^2 + ay + b)(y^2 + cy + d)$ Expanding and comparing coefficients gives $(y^2 - y-8)(y^2 + 2y - 4) .$ Using the quadratic formula for $y> 0$ only gives the other two roots: $y = \sqrt5 - 1,\frac{\sqrt{33} + 1}2$ . $$ Back-substitution gives $x = 4, \, 2 + 2\sqrt5, \, -\frac{\sqrt{33} + 1}2 .$

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If we're content with just finding the number of roots by plotting simple graphs, then consider this approach: $$ Let's rearrange the given equation: $\begin{array} { r c l} x^2 \left(1 - \frac1{\sqrt{8-x}} \right) &=&8 \\ \frac1{\sqrt{8-x}} &=&1 - \frac8{x^2} \\ \sqrt{8-x} &=& 1 + \frac{8}{x^2-8} \\ \end{array}$ Both $g (x) := \sqrt{8-x}$ and $h(x) :=1 + \frac8{x^2-8}$ are rather simple functions to plot. $$ Plotting these graphs on Desmos shows 3 intersection points: And because we didn't square the original equation, we did not introduce any extraneous roots. Indeed, there are $\boxed3$ roots.

Given that $x^2(\sqrt{8-x}-1) = 8\sqrt{8-x}$ , $\implies \sqrt{8-x} = \dfrac {x^2}{x^2-8}$ . Then

$\begin{aligned} x^2(\sqrt{8-x}-1) & = 8\sqrt{8-x} & \small \blue{\text{Multiply both sides by }\sqrt{8-x}} \\ x^2(8-x) - x^2\sqrt{8-x} & = 8(8-x) \\ 8x^2 - x^3 - \frac {x^4}{x^2-8} & = 64 - 8x \\ 8x^4 - 64x^2 - x^5 + 8x^3 - x^4 & = 64x^2 - 512 - 8x^3 + 64x \\ x^5 - 7x^4 - 16x^3 + 128x^2 + 64x - 512 & = 0 \\ (x-4)(x^4-3x^3-28x^2+16x+128) & = 0 \\ (x-4)(x^2+x-8)(x^2 - 4x-16) & = 0 \\ \implies x & = 4, \frac {\pm \sqrt{33}-1}2, \pm 2\sqrt 5 + 2 \end{aligned}$

Substituting the values of $x$ in the original equation, we find only $\boxed 3$ roots $4$ , $-\dfrac {\sqrt{33}+1}2$ , and $2\sqrt 5 +2$ .