## College Algebra (10th Edition)

$\displaystyle \{-\frac{1}{4},\frac{3}{4}\}$
$x^{2}-\displaystyle \frac{1}{2}x-\frac{3}{16}=0$ multiply with the LCD, 16 $16x^{2}-8x-3=0$ The first two terms can be recognized as $(4x)^{2}-2\cdot(4x)\cdot(1)$ The third term missing for a perfect square is $(1)^{2}$, so we rewrite the equation $((4x)^{2}-2\cdot(4x)\cdot(1)+1^{2})-1^{2}-3^{2}=0$ $(4x-1)^{2}-4=0$ Now, with $4=2^{2}$, the LHS is a difference of squares, which factors to $[(4x-1)+2][(4x-1)-2]=0$ $(4x+1)(4x-3)=0$ so either $4x+1=0\quad$or$\quad 4x-3=0$ $x=-\displaystyle \frac{1}{4}\quad$or$\displaystyle \quad x=\frac{3}{4}.$ Solution set = $\displaystyle \{-\frac{1}{4},\frac{3}{4}\}$