Question

The square The square root of 4x2+20x+25 is

Answer 1

Answer:

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)

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

= ±(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)2 = a2 + 2ab + b2

= ±(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)2 = a2 + 2ab + b2Square the value ( 2x-5)2= (2x)2 - (2)(2x)(5) + (5)2 = 4x2 -20x + 25

= ±(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)2 = a2 + 2ab + b2Square the value ( 2x-5)2= (2x)2 - (2)(2x)(5) + (5)2 = 4x2 -20x + 25Square the value -(2x-5)2 = (-2x+5)2 = (-2x)2+(2)(-2x)(5) + (5)2 = 4x2 -20x + 25

= ±(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)2 = a2 + 2ab + b2Square the value ( 2x-5)2= (2x)2 - (2)(2x)(5) + (5)2 = 4x2 -20x + 25Square the value -(2x-5)2 = (-2x+5)2 = (-2x)2+(2)(-2x)(5) + (5)2 = 4x2 -20x + 25Since both solutions lead to the same original polynomial under the radical, both solutions (2x-5)

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

Answer 2

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