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The roots of the equation $(q-r) x^{2}+(r-p) x+(p-q)=0$ are
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Verified Answer
The correct answer is:
$(\mathrm{p}-\mathrm{q}) /(\mathrm{q}-\mathrm{r}), \mathrm{j}$
Given equation, $(q-r) x^{2}+(r-p) x+(p-q)=0$
On observing the equation, it is clear that 1 is root of equation. If $x=1$, then $q-r+r-p+p-q=0$
1 is one root of given equation. Since, the given equation is quadratic equation, we
know that product of roots is $\frac{\mathrm{c}}{\mathrm{a}}$
Let the second root be $\alpha$.
$\therefore \quad(1)(\alpha)=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}} \Rightarrow \alpha=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}}$
On observing the equation, it is clear that 1 is root of equation. If $x=1$, then $q-r+r-p+p-q=0$
1 is one root of given equation. Since, the given equation is quadratic equation, we
know that product of roots is $\frac{\mathrm{c}}{\mathrm{a}}$
Let the second root be $\alpha$.
$\therefore \quad(1)(\alpha)=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}} \Rightarrow \alpha=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}}$
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