## Calculus (3rd Edition)

a) x=$\frac{-1}{4}$ ; 1 b) x = 1 ± $\sqrt 2$
a) f(x) = 4$x^{2}$ - 3x - 1 (4x+1)(x-1)=0 4x+1 = 0 ; x-1 = 0 x=$\frac{-1}{4}$ ; 1 b) f(x) = $x^{2}$ - 2x - 1 $\frac{-(-2)±\sqrt(-2)^{2}-(4*1*-1)}{2*1}$