A) \[p=q\]
B) \[{{q}^{2}}=pr\]
C) \[{{p}^{2}}=qr\]
D) \[{{r}^{2}}=pr\]
Correct Answer: B
Solution :
| Consider both equations |
| \[p{{x}^{2}}+2qx+r=0\] ... (i) |
| and \[q{{x}^{2}}-2\sqrt{pr}\,.\,x+q=0\] ... (ii) |
| Since, both the equations are quadratic and have real roots, therefore from equation (1), we have |
| \[\therefore \,\,\,\,\,4{{q}^{2}}-4pr\ge 0\] (using discriminant) |
| \[\Rightarrow \,\, {{q}^{2}}\ge pr\] ... (iii) |
| and from second equation \[4pr - 4{{q}^{2}} \ge 0\] |
| \[\Rightarrow \,\,\, pr\ge {{q}^{2}}\] ... (iv) |
| From eqs. (iii) and (iv) we get \[{{\operatorname{q}}^{2}}= pr.\] |
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