Answer
a. $x=\frac{1\pm \sqrt 5}{2},$
b. $x=-2$ or $x=3$
c. $x=-\frac{5}{2},$
d. $\frac{1- \sqrt 5}{2}
Work Step by Step
$f(x)=-x^2-x+1,$
$g(x)=-x^2+x+6,$
a. $-x^2-x+1=0,$
Solving the quadratic equation using the quadratic formula,
$x=\frac{-b\pm \sqrt {b^2-4ac}}{2a},$
In this case,
$x=\frac{1\pm \sqrt {(-1)^2-4(-1)(1)}}{2(-1)}=\frac{1\pm \sqrt 5}{2},$
b. $-x^2+x+6=0,$
$-x^2-2x+3x+6=0,$
$-x(x+2)+3(x+2)=0,$
$(-x+3)(x+2)=0,$
$x=-2$ or $x=3$
c. $-x^2-x+1=-x^2+x+6,$
$-2x-5=0,$
$x=-\frac{5}{2},$
d. $-x^2-x+1>0,$
$\frac{1- \sqrt 5}{2}-x^2+x+6,$
$-2x-5>0,$
$x<-\frac{5}{2}$
g. $-x^2-x+1\geq1,$
$-x^2-x\geq0,$
$-x(x+1)\geq0,$
$-1\leq x\leq 0$
