A) 0 only
B) 0 and 3
C) \[-\,\frac{1}{2}\] and 3
D) 0 and 7
Correct Answer: D
Solution :
[d] \[(x-\,3)\,\,(2x+1)=0\] \[\Rightarrow (x-\,3)=0\,\,or\,\,(2x+1)=0\] \[when\,\,x-\,3=0\,\,then\,\,x=3\] \[when\,\,2x+1=0\,\,then\,\,x=-\,\frac{1}{2}\] If x=3, then (2x+1) =7 And 2x+1=0 When x=\[-\,\frac{1}{2}\] Possible value of (2x + 1) are 0 and 7.You need to login to perform this action.
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