A) \[2b=a+c\]
B) \[{{b}^{2}}=ac\]
C) \[b=\frac{2ac}{a+c}\]
D) \[b=ac\]
Correct Answer: B
Solution :
Since roots of the given equation are equal \[\therefore \]\[D=0\] \[\Rightarrow \] \[{{(-2b(a+c))}^{2}}-4({{a}^{2}}+{{b}^{2}})({{b}^{2}}+{{c}^{2}})=0\] \[\Rightarrow \] \[4{{b}^{2}}({{a}^{2}}+{{c}^{2}}+2ac)-4({{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{4}}\] \[+{{b}^{2}}{{c}^{2}})=0\] \[\Rightarrow \]\[{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+2a{{b}^{2}}c-{{a}^{2}}{{b}^{2}}-{{a}^{2}}{{c}^{2}}-{{b}^{4}}-{{b}^{2}}{{c}^{2}}=0\] \[\Rightarrow \]\[2a{{b}^{2}}c-{{a}^{2}}{{c}^{2}}-{{b}^{4}}=0\]\[\Rightarrow \]\[{{b}^{4}}+{{a}^{2}}{{c}^{2}}-2a{{b}^{2}}c=0\] \[\Rightarrow \]\[{{({{b}^{2}}-ac)}^{2}}=0\] \[\Rightarrow \] \[{{b}^{2}}=ac\]You need to login to perform this action.
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