How do you solve #25x^2-64#?

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Answer 1 · 2015-06-28T08:31:48

This is a difference of squares:

#25x^2-64 = (5x)^2 - 8^2 = (5x-8)(5x+8)#

So the roots of #25x^2-64 = 0# are #8/5# and #-8/5#

Any difference of squares is of the form #a^2-b^2 = (a-b)(a+b)#

In our case #a = 5x# and #b=8#, giving:

#25x^2-64 = (5x)^2 - 8^2#

#=a^2 - b^2#

#=(a-b)(a+b)#

#= (5x-8)(5x+8)#