Question

In the given problem, solve and graph. Express answers in both inequality and interval notaion.$2x+2>x^2$

Answer 1

$2x+2>x^2$

$x^2<2x+2$

$x^2-2x-2<0$

$x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4\cdot 1(-2)}}{2\cdot 1}:1\pm \sqrt{3}$

$(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))<0$

	x< $1-\sqrt{3}$	x=$1-\sqrt{3}$	$1-\sqrt{3}<x<1+\sqrt{3}$	x=$1+\sqrt{3}$	x>$1+\sqrt{3}$
x-$(1+\sqrt{3})$	-	-	-	0	+
x-$(1-\sqrt{3})$	-	0	+	+	+
(x-$(1+\sqrt{3}$))(x-$(1-\sqrt{3}$))	+	0	-	0	+

$\left[\begin{array}{cc}Solution:& 1-\sqrt{3}<x<1+\sqrt{3}\\ Decimal:& -0.73205\cdots <x<2.73205\dots \\ IntervalNotation:& (1-\sqrt{3},1+\sqrt{3})\end{array}\right]$