If $\sin\theta$ and $\cos\theta$ are the roots of the quadratic equation $px^2+qx+r=0$ $(p \neq 0)$, then which of the following relation holds good?

Since $\sin\theta$ and $\cos\theta$ are roots of $px^2 + qx + r = 0$, by Vieta's formulas:

$\sin\theta + \cos\theta = -\dfrac{q}{p}$

$\sin\theta \cdot \cos\theta = \dfrac{r}{p}$

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Squaring the first equation:

$\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = \dfrac{q^2}{p^2}$

Since $\sin^2\theta + \cos^2\theta = 1$:

$1 + 2\cdot\dfrac{r}{p} = \dfrac{q^2}{p^2}$

$\dfrac{p + 2r}{p} = \dfrac{q^2}{p^2}$

$p^2 + 2pr = q^2$

$q^2 - p^2 = 2pr$

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Comparing with the given options:

Option 2: $q^2 + p^2 + 2pr = 0$ — does not match.

Option 3: $p^2 - q^2 = 2pr$ — the sign is opposite to $q^2 - p^2 = 2pr$.

Option 4: $(q - p)^2 = 2pr$ expands to $q^2 - 2pq + p^2 = 2pr$ — has an extra $-2pq$ term.

None of the listed relations match $q^2 - p^2 = 2pr$, so the answer is None of these.