How do you solve the inequality #2x^2-4x-1>0# and write your answer in interval notation?

1 Answer
Mar 6, 2017

Answer:

The solution is #x in ]-oo,(1-sqrt6/2)[ uu ] (1+sqrt6/2), +oo[#

Explanation:

First, we must solve the equation

#2x^2-4x-1=0#

The discriminant is

#Delta=b^2-4ac=(-4)^2-4(2)(-1)=16+8=24#

As, #Delta>0#, there are 2 real roots

#x=(-b+-sqrt(Delta))/(2a)#

#x_1=(4-sqrt24)/4=(4-2sqrt6)/4=1-sqrt6/2=-0.225#

#x_2=(4+sqrt24)/4=(4+2sqrt6)/4=1+sqrt6/2=2.225#

Let #f(x)=2x^2-4x-1#

Now, we build the sign chart

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

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

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

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

Therefore,

#f(x)>0#, when #x in ]-oo,(1-sqrt6/2)[ uu ] (1+sqrt6/2), +oo[#