How do you solve #25-x^2>=0#?

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Answer 1 · 2018-04-10T12:32:58

Solution: # -5 <= x <= 5 or [-5,5] #

#25-x^2 >= 0 or (5+x)(5-x)>=0 #

Critical points are

#5+x=0 or x= -5 and 5-x=0 or x= 5 #

#f(x)=0# when # x=-5 and x=5#

Sign chart:

When #x< -5# sign of #(5+x)(5-x) # is # (-) * (+) = (-) ; < 0#

When # -5 < x < 5 # sign of #(5+x)(5-x) # is # (+) * (+) = (+) ; > 0#

When #x> 5# sign of #(5+x)(5-x) # is # (+) * (-) = (-) ; < 0#

Solution: # -5 <= x <= 5 or [-5,5] # [Ans]

Answer 2 · 2018-04-10T12:36:59

#color(blue)([ -5 ,5]#

#25 -x^2 >= 0#

Factor #LHS#:

#-(5+x)(5-x)>=0#

#(5+x)(5-x)<=0#

For:

#5+x<=0#

#x<=-5#

For:

#5-x<=0#

#x>=5#

Solutions:

#color(blue)([ -5 ,5]#