How do you solve #x^2-2x-4>0# using a sign chart?

1 Answer
Jul 14, 2017

Answer:

The solution is #x in (-oo, (1-sqrt5)) uu((1+sqrt5), +oo)#

Explanation:

We must first find the roots of the quadratic equations

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

The discriminant is

#Delta=b^2-4ac=(-2)^2-4*(1)*(-4)=20#

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

The roots are

#x=(-b+-sqrtDelta)/(2a)#

#x_1=(2-sqrt20)/2=1-sqrt5#

#x_2=(2+sqrt20)/2=1+sqrt5#

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

We can 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)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##(x-x_2)##color(white)(aaaaa)##-##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 (-oo, (1-sqrt5)) uu((1+sqrt5), +oo)#