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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 4 · Set 1 · Q21
Verify that roots of the quadratic equation ( p - q ) x^2+( q - r ) x+( r - p )=0 are equal when q + r =2 p.
Verify that roots of the quadratic equation $\displaystyle (\mathrm{p}-\mathrm{q}) x^{2}+(\mathrm{q}-\mathrm{r}) x+(\mathrm{r}-\mathrm{p})=0$ are equal when $\displaystyle \mathrm{q}+\mathrm{r}=2 \mathrm{p}$.
Marking-scheme solution
Discriminant \(\displaystyle (\mathrm{D})=(\mathrm{q}-\mathrm{r})^{2}-4(\mathrm{p}-\mathrm{q})(\mathrm{r}-\mathrm{p})\)
\[=(\mathrm{q}+\mathrm{r}-2 \mathrm{p})^{2}
\]
Substituting, \(\displaystyle \mathrm{q}+\mathrm{r}=2 \mathrm{p}\)
\[\Rightarrow \mathrm{D}=(2 \mathrm{p}-2 \mathrm{p})^{2}=0
\]
∴ Roots of the given equation are equal.
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