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:

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

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

$2x-5=0$

$x=5/2$

How do you solve 4x² - 20x + 25 = 04x² - 20x + 25 = 0?