Mathematics — Standard (041) — CBSE Class 10 board question

Here, $a = (p-q)$, $b = (q-r)$, $c = (r-p)$.

Discriminant $D = b^2 - 4ac$

$= (q-r)^2 - 4(p-q)(r-p)$

Given $q + r = 2p$, so $q - r = (q-r)$ and we can write $q - p = p - r$ (since $q + r = 2p \Rightarrow q - p = p - r$).

Let $q - p = p - r = k$, so $q - r = 2k$, $p - q = -k$, $r - p = -k$.

$D = (2k)^2 - 4(-k)(-k) = 4k^2 - 4k^2 = 0$

Since $D = 0$, the roots are equal.

Source: Chapter 4, Section 4.4 – Nature of Roots

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• The examiner wants you to compute the discriminant $D = b^2 - 4ac$ and show $D = 0$ using the given condition $q + r = 2p$.
• The substitution $q - p = p - r$ (derived from $q + r = 2p$) is the key step — write it clearly.
• Conclude explicitly: "Since $D = 0$, roots are equal." Don't skip this line — it fetches marks.
• Avoid expanding everything in full; using the substitution $k$ keeps work neat and error-free.