Let $r$ and $c$ be real numbers, if $r$ and $-r$ are roots of $5 x^{3}+c x^{2}-10 x+9=0$, then $c$ equals

We have a cubic equation: $5x^3 + cx^2 - 10x + 9 = 0$

We're told that $r$ and $-r$ are roots of this equation. Since they're opposites of each other, this gives us a special structure we can exploit.

A cubic equation has exactly 3 roots. Since we know 2 of them ($r$ and $-r$), there must be a third root. Let's call it $p$.

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For any cubic equation $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$, Vieta's formulas tell us:

Sum of roots: $\alpha + \beta + \gamma = -\tfrac{b}{a}$

Sum of products taken two at a time: $\alpha\beta + \alpha\gamma + \beta\gamma = \tfrac{c}{a}$

Product of all roots: $\alpha\beta\gamma = -\tfrac{d}{a}$

These relationships come from expanding $(x-\alpha)(x-\beta)(x-\gamma)$ and comparing coefficients.

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Our equation: $5x^3 + cx^2 - 10x + 9 = 0$

Our roots: $r, -r, p$ (where $p$ is the unknown third root)

Sum of roots:

$r + (-r) + p = -\dfrac{c}{5}$

$0 + p = -\dfrac{c}{5}$

$p = -\dfrac{c}{5}$ ... (equation 1)

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Sum of products taken two at a time:

$r(-r) + r(p) + (-r)(p) = \dfrac{-10}{5}$

$-r^2 + rp - rp = -2$

$-r^2 + 0 = -2$ (since $+rp - rp = 0$)

$-r^2 = -2$

$r^2 = 2$

$r = \pm\sqrt{2}$

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Product of all roots:

$r(-r)(p) = -\dfrac{9}{5}$

$-r^2 \cdot p = -\dfrac{9}{5}$

Since $r^2 = 2$:

$-2p = -\dfrac{9}{5}$

$p = \dfrac{9}{10}$ ... (equation 2)

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From equations 1 and 2:

$p = -\dfrac{c}{5}$ and $p = \dfrac{9}{10}$

Therefore: $\dfrac{9}{10} = -\dfrac{c}{5}$

$9 \times 5 = -c \times 10$

$45 = -10c$

$c = -\dfrac{45}{10} = \boxed{-\dfrac{9}{2}}$