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

1 Answer
Sep 20, 2016

Answer:

Either #y < -2# or #y > 2#

Explanation:

As #y^2-4>0#, factorizing we have #(y+2)(y-2)>0#

From this we know that the product #(y+2)(y-2)# is positive. It is apparent that sign of binomials #(y+2)# and #y-2# will change around the values #-2# and #2# respectively. In sign chart we divide the real number line using these values, i.e. below #-2#, between #-2# and #2# and above #2# and see how the sign of #y^2-4# changes.

Sign Chart

#color(white)(XXXXXXXXXXX)-2color(white)(XXXXX)2#

#(y+2)color(white)(XXX)-ive color(white)(XXXX)+ive color(white)(XXXX)+ive#

#(y-2)color(white)(XXX)-ive color(white)(XXXX)-ive color(white)(XXXX)+ive#

#(y^2-4)color(white)(XXX)+ive color(white)(XXX)-ive color(white)(XXXX)+ive#

It is observed that #y^2-4 > 0# when either #y < -2# or #y > 2#, which is the solution for the inequality.