How do you solve #x^2-1>=4x# using a sign chart?

1 Answer
Dec 14, 2016

Answer:

The answer is #x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[#

Explanation:

Let's rewrite the equation

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

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

We need the values of #x#, when #f(x)=0#

that is, #x^2-4x-1=0#

Let's calculate #Delta=16-4*1*-1=20#

#Delta>0#. there are 2 real solutions

#x_1=(4+sqrt20)/2=(4+2sqrt5)/2=2+sqrt5#

#x_2=(4-sqrt20)/2=(4-2sqrt5)/2=2-sqrt5#

Now, we can do our sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-oo##color(white)(aaaa)##(2-sqrt5)##color(white)(aaaa)##(2+sqrt5)##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-2+sqrt5##color(white)(aaaaa)##-##color(white)(aaaaaaaaaa)##+##color(white)(aaaaaaaaa)##+#

#color(white)(aaaa)##x-2-sqrt5##color(white)(aaaaa)##-##color(white)(aaaaaaaaaa)##-##color(white)(aaaaaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaaa)##+##color(white)(aaaaaaaaaa)##-##color(white)(aaaaaaaaa)##+#

Therefore,

#f(x>=0)# when #x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[#

graph{x^2-4x-1 [-11.25, 11.25, -5.625, 5.625]}