How do you solve #x^2-16/25=0#?

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Answer 1 · 2016-10-25T19:49:38

#x = +4/5, x = -4/5#

There are two different methods we can use.

#x^2 -16/25 =0#

#(x+4/5)(x-4/5)=0#

Either factor could be equal to 0.

#x + 4/5 = 0 " "rarr x = -4/5#

#x - 4/5 = 0 " "rarr x = +4/5#

#2.# Find the square root.

#x^2 = 16/25#

#x = +-sqrt(16/25)#

#x = +4/5, x = -4/5#