Question:
The roots of the equation \((q – r)x^2 + (r – p)x + (p – q) = 0\) are
समीकरण \((q – r)x^2 + (r – p)x + (p – q) = 0\) के मूल (roots) हैं:
Options:
- \(\dfrac{p-q}{q-r}\) and 1
- \(\dfrac{q-r}{p-q}\) and 1
- \(\dfrac{r-p}{q-r}\) and \(\dfrac{1}{2}\)
- \(\dfrac{r-p}{p-q}\) and \(\dfrac{1}{2}\)
- \(\dfrac{p-q}{q-r}\) और 1
- \(\dfrac{q-r}{p-q}\) और 1
- \(\dfrac{r-p}{q-r}\) और \(\dfrac{1}{2}\)
- \(\dfrac{r-p}{p-q}\) और \(\dfrac{1}{2}\)
Correct Answer: a)
Explanation:
Sum of coefficients of \((q-r)x^2 + (r-p)x + (p-q) = 0\) is \((q-r) + (r-p) + (p-q) = 0\).
Hence \(x = 1\) is a root.
Product of roots is \(1 \cdot \beta = \dfrac{p-q}{q-r} \implies \beta = \dfrac{p-q}{q-r}\).
Thus roots are \(1\) and \(\dfrac{p-q}{q-r}\).
समीकरण \((q-r)x^2 + (r-p)x + (p-q) = 0\) के गुणांकों का योग \((q-r) + (r-p) + (p-q) = 0\) है।
अतः एक मूल सदैव \(1\) होगा।
मूलों का गुणनफल \(1 \cdot \beta = \dfrac{p-q}{q-r} \implies \beta = \dfrac{p-q}{q-r}\)।
अतः समीकरण के मूल \(1\) और \(\dfrac{p-q}{q-r}\) हैं।