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