A) \[\frac{p}{r}=\frac{q}{s}\]
B) \[2h=\left[ \frac{p}{q}+\frac{r}{s} \right]\]
C) \[{{p}^{2}}-4q={{r}^{2}}-4s\]
D) \[p{{r}^{2}}=q{{s}^{2}}\]
Correct Answer: C
Solution :
\[\alpha +\beta =-p,\,\alpha \beta =q\] \[\alpha +\beta +2h=-r,\] \[(\alpha +h)(\beta +h)=s\] \[-p+2h=-r\Rightarrow h=\frac{p-r}{2}\] ?..(i) Now, \[\alpha \beta +h\,(\alpha +\beta )+{{h}^{2}}=s\] \[\Rightarrow q+h(-p)+{{h}^{2}}=s\] \[\Rightarrow \,\,q+\left( \frac{p-r}{2} \right)\,(-p)+{{\left( \frac{p-r}{2} \right)}^{2}}=s\] \[\Rightarrow q-\frac{({{p}^{2}}-pr)}{2}+\frac{{{p}^{2}}+{{r}^{2}}-2pr}{4}=s\] \[\Rightarrow 4q-2{{p}^{2}}+2pr+{{p}^{2}}+{{r}^{2}}-2pr=4s\] \[\Rightarrow 4q-{{p}^{2}}+{{r}^{2}}-4s=0\] \[\Rightarrow {{r}^{2}}-4s={{p}^{2}}-4q\].
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