Answer :
Answer:
Required side of the squire is [tex]2.4721[/tex], correct upto four desimal places.
Step-by-step explanation:
Given two inequalities are,
[tex]y<-\frac{1}{2}x^2+2\hfill (1)[/tex]
[tex]y>\frac{1}{2}x^2-2\hfill (2)[/tex]
which are reflexions of each other across the [tex]x-[/tex]axis. And the square is inscribed within it. Since by symmetry, vertices of the square lies on the lines,
[tex]y=x\hfill (3)[/tex]
and [tex]y=-x\hfill (4)[/tex]. We only have to find length of any one side of the squire.
To find [tex]P[/tex], substitute [tex](3)[/tex] in [tex](1)[/tex] we get,
[tex] x^2+2x-4=0[/tex]
[tex]x=\frac{-2\pm\sqrt{4+16}}{2}=-1\pm \sqrt{5}[/tex]
Since [tex]x>0, x=-1+\sqrt{5}[/tex] and from [tex](3)[/tex], [tex]y=-1+\sqrt{5}[/tex], and thus,
[tex]P=(-1+\sqrt{5}, -1+\sqrt{5})[/tex]
Similarly by substitute [tex](4)[/tex] and [tex](2)[/tex] we will get,
[tex]R=(-1+\sqrt{5}, -1+\sqrt{5})[/tex]
And thus, [tex]\bar{PR}=s=|2\sqrt{5}-2|=2.4721[/tex], correct upto four desimal places.