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

1 Answer
Mar 29, 2016

Answer:

The solutions are:

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

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

Explanation:

#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#