A) \[(-1+\sqrt{6})a\]
B) b)\[(\sqrt{6}-1)a\]
C) \[a\]
D) none of these
Correct Answer: B
Solution :
Given,\[{{x}^{2}}-2a|x-a|-3{{a}^{2}}=0\] Here \[a\le 0\] We know that \[|x-a|\,\,=\left\{ \begin{align} & x-a,\,\,x>a \\ & -(x-a),\,\,x<a \\ \end{align} \right.\] Thus \[{{x}^{2}}-2a|x-a|-3{{a}^{2}}=0\] gives two cases Case I: When \[x>a\] \[\therefore \] \[{{x}^{2}}-2a(x-a)-3{{a}^{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}-2ax-{{a}^{2}}=0\Rightarrow x=a\pm \sqrt{2}a\] Now, for\[x\ge a,\,\,a<0\Rightarrow x=a(1-\sqrt{2})\] ... (i) \[(\because \,\,x=a(1+\sqrt{2})<a)\] Case II: When\[x<a\] \[\therefore \] \[{{x}^{2}}+2a(x-a)-3{{a}^{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}+3ax-5{{a}^{2}}=0\]\[\Rightarrow \]\[x=-a\pm \sqrt{6}a\] Now, for \[x<a,\,\,a<0\Rightarrow x=a(\sqrt{6}-1)\] ... (ii) \[(\because x=-a\cdot (1+\sqrt{6})>a)\] From (i) and (ii), \[x=\{a(1-\sqrt{2}),\,\,a(\sqrt{6}-1)\}\]
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