How do you solve #(4x - 3)^{2} = 25x^{2}#?

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Answer 1 · 2016-09-24T19:52:13

#x = -3 or x = 1/3#

We could get rid of the brackets by expanding the left hand side, but let's rather get rid of the square!

Square root both sides:

#4x-3 = +-sqrt(25x^2)#

Now:either

#4x -3 = 5x, " "rarr -3 = x#

OR

#4x-3 = -5x" " rarr 9x= 3" "rarr x = 1/3#