How do you solve #4y^2 = 25#?

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Answer 1 · 2016-03-29T16:26:13

The solutions are:

#color(blue)(y=-5/2#

#color(blue)(y=5/2#

#4y^2 = 25#

#4y^2 - 25 = 0#

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

Applying the below mentioned property to the expression:
#color(blue)(a^2- b^2 = (a+b)(a-b)#

#(2y)^2 - 5^2 = color(blue)((2y+5)(2y-5)#

Equating each of the factors to zero , we obtain the solution:

#2y + 5 = 0, color(blue)(y=-5/2#

#2y- 5 = 0, color(blue)(y=5/2#