How do you solve #4x² - 20x + 25 = 0#?

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Answer 1 · 2015-07-01T12:48:28

I would solve by factoring.

#4x^2-20x+25 = 0#

Look at #4x^2-20x+25#

the first term is a perfect square: #4x^2 = (2x)^2#

the last term is a perfect square: #25 = 5^2#

If we double the product f the things we squared, we get:

#2*(2x)(5) = 20x#, which is the absolute value of the middle term.

#(ax-b)^2 = a^2x^2-2abx+b^2#, so we can factor:

#4x^2-20x+25 = (2x-5)^2#

With practice, there is no need to write the stuff up to this point, we can write:

#4x^2-20x+25 = 0#

#(2x-5)^2=0#

#2x-5=0#

#x=5/2#