Answer
a. see explanations.
b. $0.8555986772$
Work Step by Step
a. Letting $y=x^4+2x^2-2$, we have $y'=4x^3+4x$. Within $[0,1]$, we have $y'\gt0$; thus the function increases over the interval. Test end points, $f(0)=-2\lt0, f(1)=1+2-2=1\gt0$; thus there is only one solution to the equation $y=0$.
b. We can solve the equation graphically as shown to get $x=0.856$ in $[0,1]$. We can also solve it algebraically by letting $z=x^2$. The equation becomes $z^2+2z-2=$, which gives $z=\frac{-2\pm\sqrt {4+8}}{2}=-1\pm\sqrt 3 $. Choosing $x^2=z=\sqrt 3-1$, we have, in $[0,1]$, $x=\sqrt {\sqrt 3-1}\approx0.8555986772$.
