How do you solve #2x+2>x^2# using a sign chart?

1 Answer
May 23, 2017

Answer:

The solution is # x in (1-sqrt3, 1+sqrt3)#

Explanation:

Let's rewrite the equation

#2x+2>x^2#

#x^2-2x-2<0#

The roots of the equation

#x^2-2x-2=0#

are #x_1# and #x_2#

#x_2=(2+sqrt(2^2-(4*1*-2)))/2=(2+sqrt12)/2=(2+2sqrt3)/2=1+sqrt3#

#x_1=(2-sqrt(2^2-(4*1*-2)))/2=(2-sqrt12)/2=(2-2sqrt3)/2=1-sqrt3#

Let #f(x)=x^2-2x-2#

We build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##x_2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-x_1##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-x_2##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<0#, when #x in (x_1,x_2)#