Question 127 of 161MediumMCQ
Let S={x∈R:x≥0 and 2∣x−3∣+x(x−6)+6=0}. Then S:
Solutions · 1
Official solution
Given,
\[2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0\]
Case 1 :
When \[\sqrt x - 3 \ge 0,\] then equation becomes
\[2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0\]
\[ \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0\]
\[ \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0\]
\[ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0\]
\[\therefore\,\,\,\] \[\sqrt x = 0,4\]
but as \[\sqrt x - 3 \ge 0\] or \[\sqrt x \ge 3\] then \[\sqrt x \ne 0\]
\[\therefore\,\,\,\] \[\sqrt x = 4\] value is possible.
Case 2 :
When \[\sqrt x - 3 < 0\] or \[\sqrt x < 3.\] There equation becomes
\[ - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0\]
\[ \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0\]
\[ \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0\]
\[ \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0\]
\[ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0\]
\[ \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0\]
\[\therefore\,\,\,\] \[\sqrt x = 2,6\]
as \[\sqrt x < 3\] so \[\sqrt x \ne 6\]
\[\therefore\,\,\,\] \[\sqrt x = 2\] is possible.
So, total possible value of \[\sqrt x = 2,4\]
or for x possible values are 4, 16.
\[\therefore\,\,\,\] Set S contains exactly two elements 4 and 16.
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