A) \[\Delta \ne 0\]
B) \[b\Delta =0\]
C) \[cb\ne 0\]
D) \[c\Delta =0\]
Correct Answer: D
Solution :
[d] \[\frac{-b}{a},\frac{{{b}^{2}}-2ac}{{{a}^{2}}},\frac{-{{b}^{3}}+3abc}{{{a}^{3}}}\] are in G.P. \[{{\left( {{b}^{2}}-2ac \right)}^{2}}=\left( -{{b}^{3}}+3abc \right)\left( -b \right)\] \[{{b}^{4}}+4{{a}^{2}}{{\text{c}}^{2}}-4a{{b}^{2}}c={{b}^{4}}-3a{{b}^{2}}c\] \[a{{b}^{2}}c-4{{a}^{2}}{{c}^{2}}=0\] a.c. \[({{b}^{2}}-4ac)=0\] \[\Rightarrow c.\Delta =0\]You need to login to perform this action.
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