A) \[a+b+4=0\]
B) \[a+b-4=0\]
C) \[a-b-4=0\]
D) \[a-b+4=0\]
Correct Answer: A
Solution :
Let \[{{\alpha }_{1}},{{\beta }_{1}}\] are the roots of the eqn \[{{x}^{2}}+ax+b=0\]\[\Rightarrow x=\frac{-a\pm \sqrt{{{a}^{2}}-4b}}{2}\] Þ \[{{\alpha }_{1}}=\frac{-a+\sqrt{{{a}^{2}}-4b}}{2},{{\beta }_{1}}=\frac{-a-\sqrt{{{a}^{2}}-4b}}{2}\] and \[{{\alpha }_{2}},{{\beta }_{2}}\] are the roots of the equation \[{{x}^{2}}+bx+a=0\] So, \[{{\alpha }_{2}}=\frac{-b+\sqrt{{{b}^{2}}-4a}}{2},\]\[{{\beta }_{2}}=\frac{-b-\sqrt{{{b}^{2}}-4a}}{2}\] Now \[{{\alpha }_{1}}-{{\beta }_{1}}=\sqrt{{{a}^{2}}-4b}\]; \[{{\alpha }_{2}}-{{\beta }_{2}}=\sqrt{{{b}^{2}}-4a}\] Given, \[{{\alpha }_{1}}-{{\beta }_{1}}={{\alpha }_{2}}-{{\beta }_{2}}\] \[\Rightarrow \sqrt{{{a}^{2}}-4b}=\sqrt{{{b}^{2}}-4a}\] \[\Rightarrow {{a}^{2}}-{{b}^{2}}=-\,4\,(a-b)\] \[\Rightarrow a+b+4=0\].
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