Question

If $x(2x+1)=0$ and $(x+\frac{1}{2})(2x-3)=0$, then $x=$

• A. $-3$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\frac{1}{2}$
• E. $\frac{3}{2}$

Answer 1

The correct option is B $-\frac{1}{2}$

$x(2x+1)=0x(x+\frac{1}{2})=0x=0,\frac{-1}{2}\&(x+\frac{1}{2})(2x-3)=0(x+\frac{1}{2})(x-\frac{3}{2})=0x=\frac{-1}{2},\frac{3}{2}\therefore x\in \frac{-1}{2}$