A) \[(-1+\sqrt{6})a\]
B) \[(\sqrt{6}-1)a\]
C) a
D) None of these
Correct Answer: A
Solution :
For \[x\ge a,\] , the equation becomes \[{{x}^{2}}-2a(x-a)-3{{a}^{2}}=0\Rightarrow x=(1+\sqrt{2})a,(1-\sqrt{2})a\]for \[x\le a,\], the equation becomes \[{{x}^{2}}-2a[(x-a)]-3{{a}^{2}}=0\Rightarrow {{x}^{2}}+2ax-5{{a}^{2}}=0\]\[\Rightarrow x=-(1+\sqrt{6})a,(-1+\sqrt{6})a\] This shows \[(-1+\sqrt{{}})\]a is one of the roots.
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