MCQ
સમીકરણ $(p - q)x^2 + (q - r)x + (r - p) = 0$ ના બીજ.......છે.
- A$\frac{{p - q}}{{r - p}},1$
- B$\frac{{q - r}}{{p - q}},1$
- ✓$\frac{{r - p}}{{p - q}},1$
- D$1,\frac{{q - r}}{{p - q}}$
$x = \frac{{(r - q) \pm \sqrt {{{(q - r)}^2} - 4(r - p)(p - q)} }}{{2(p - q)}}$
$ \Rightarrow \,x = \frac{{(r - q) \pm (q + r - 2p)}}{{2(p - q)}}$
$ \Rightarrow x = \frac{{r - p}}{{p - q}},1$
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