Question

what is 4y^2-81=0

URGENT PLEASE ANSWER. what is 4y^2-81=0

URGENT PLEASE ANSWER. @Mathematics

Answer 1

y = 9/2 or -9/2

Answer 2

thanks! u worked it out quickly! how?

Answer 3

4y^2=81

y^2= 81/4

y= 9/2; -9/2

Answer 4

it is simple based on formula \[(a^2 - b^2) = (a-b)(a+b)\]

Answer 5

in this case a = 2y and b = 9

Answer 6

\[4y^2=81, \sqrt{4y}=\sqrt{81}, 2y=9, y=9/2\]

Answer 7

it's a difference of squares. 4y^2 is (2y)^2 and 81 is 9^2 It becomes (2y-9)(2y+9) = 0
set each of the terms equal to zero. 2y-9=0 and 2y+9=0

You get y=9/2 or -9/2