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The roots of the given equation $(p-q) x^2+(q-r) x+(r-p)=0$ are :
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Verified Answer
The correct answer is:
$\frac{\mathrm{r}-\mathrm{p}}{\mathrm{p}-\mathrm{q}}, 1$
Given equation is
$(p-q) x^2+(q-r) x+(r-p)=0$
By using formula for finding the root viz: $\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$, we get
$x=\frac{(r-q) \pm \sqrt{(q-r)^2-4(r-p)(p-q)}}{2(p-q)}$
$\Rightarrow \quad x=\frac{(r-q) \pm(q+r-2 p)}{2(p-q)}=\frac{r-p}{p-q}, 1$
$(p-q) x^2+(q-r) x+(r-p)=0$
By using formula for finding the root viz: $\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$, we get
$x=\frac{(r-q) \pm \sqrt{(q-r)^2-4(r-p)(p-q)}}{2(p-q)}$
$\Rightarrow \quad x=\frac{(r-q) \pm(q+r-2 p)}{2(p-q)}=\frac{r-p}{p-q}, 1$
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