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What is the square root of 4x^2-20x+25

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2 Answers

Factor the 2nd degree polynomial under the radical √(4x2-20x+25) = √(2x-5)(2x-5)=√(2x-5)2

= ±(2x-5) and therefore there are two possible solutions; (2x-5) and -(2x-5)

Check the two possible solutions:

Knowing that  (a-b)2 = a2 - 2ab + b2  and (a+b)= a2 + 2ab + b2

Square the value ( 2x-5)2= (2x)2 - (2)(2x)(5) + (5)2 = 4x2 -20x + 25

Square the value -(2x-5)2 = (-2x+5)2 = (-2x)2+(2)(-2x)(5) + (5)2 = 4x2 -20x + 25

Since both solutions lead to the same original polynomial under the radical, both solutions (2x-5)

and -(2x-5) are valid solutions.

You can factor it:

4x2 - 20x + 25 = 22x2- 2(2)(5)x + 5= (2x - 5)2

There are two square roots, ±(2x - 5).