A) \[\sum{{{a}^{2}}={{a}^{2}}{{b}^{2}}{{c}^{2}}}\]
B) \[{{b}^{2}}+{{c}^{2}}={{a}^{2}}\]
C) \[{{b}^{2}}+{{c}^{2}}=2{{a}^{2}}\]
D) \[{{(a+b+c)}^{2}}=ab+bc+ca\]
Correct Answer: C
Solution :
(c): Trivial Solution is got by observance, put x = 1 and equation is satisfied. Hence x = 1 is a root: According to question, other root is also 1 \[\therefore \] sum of roots \[=1+1=2=\frac{-\left( {{c}^{2}}-{{b}^{2}} \right)}{\left( {{b}^{2}}-{{a}^{2}} \right)}\] \[\Rightarrow 2{{b}^{2}}-2{{a}^{2}}={{b}^{2}}-{{c}^{2}}\Rightarrow {{b}^{2}}+{{c}^{2}}=2{{a}^{2}}\] Product of roots \[=1\times 1=1=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{a}^{2}}}\Rightarrow \] again \[{{b}^{2}}+{{c}^{2}}=2{{a}^{2}}\]
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