A) \[p=q\]
B) \[{{q}^{2}}=pr\]
C) \[{{p}^{2}}=qr\]
D) \[{{r}^{2}}=pq\]
Correct Answer: B
Solution :
Equations \[p{{x}^{2}}+2qx+r=0\] and \[q{{x}^{2}}-2(\sqrt{pr})x+q=0\] have real roots, then from first \[4{{q}^{2}}-4pr\ge 0\]Þ \[{{q}^{2}}-pr\ge 0\,\,\Rightarrow {{q}^{2}}\ge pr\] .....(i) and from second \[4(pr)-4{{q}^{2}}\ge 0\](for real root ) Þ \[pr\ge {{q}^{2}}\] .....(ii) From (i) and (ii), we get result\[{{q}^{2}}=pr\].
You need to login to perform this action.
You will be redirected in
3 sec
