How do you solve #w^2-2<=0# using a sign chart?

Question

1106 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-26T12:43:38

Solution: # -sqrt2 <= w <= sqrt2 or [-sqrt2,sqrt2]#

#w^2-2 <=0 or (w+sqrt2)(w-sqrt2) <= 0 # . Critical points

are # w= sqrt2 and w = -sqrt2 # at #w = +- sqrt2 ; w^2-2=0#

Sign chart:

When #w < -sqrt2# sign of #(w+sqrt2)(w-sqrt2)# is #(-)*(-) =+ ; >0 #

When # -sqrt2 < w < sqrt2# sign of #(w+sqrt2)(w-sqrt2)#

is #(+)*(-) =- ; < 0 #

When #w > sqrt2# sign of #(w+sqrt2)(w-sqrt2)# is #(+)*(+) =+ ; >0 #

Solution: # -sqrt2 <= w <= sqrt2 or [-sqrt2,sqrt2]#

graph{x^2-2 [-12.66, 12.65, -6.33, 6.33]} [Ans]