A) \[\frac{r-p}{q-r},1\]
B) \[\frac{p-q}{q-r},1\]
C) \[\frac{p-r}{q-r},2\]
D) \[\frac{q-r}{p-q},2\]
E) \[\frac{r-p}{p-q},1\]
Correct Answer: B
Solution :
\[(q-r){{x}^{2}}+(r-p)x+p-q=0\] \[\Rightarrow \] \[{{x}^{2}}\frac{(r-p)}{q-r}x+\frac{p-q}{q-r}=0\] \[\Rightarrow \]\[{{x}^{2}}\left( \frac{r-p+q-q}{q-r} \right)x+\frac{p-q}{q-r}=0\] \[\Rightarrow \]\[{{x}^{2}}-\left( \frac{q-r}{q-r}+\frac{p-q}{q-r} \right)x+\frac{p-q}{q-r}=0\] \[\Rightarrow \]\[{{x}^{2}}-\left( 1+\frac{p-q}{q-r} \right)x+\frac{p-q}{q-r}=0\] \[\therefore \] Roots are \[1,\frac{p-q}{q-r}\]
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