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

1 Answer
Oct 25, 2016

Answer:

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

Explanation:

There are two different methods we can use.

  1. Factorise as Difference of squares

#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#