Question

1-4y^2 = 0 Solve the equation

Answer 1

EXPLANATION:

 $1-4y^2=0\\1^2-(2y)^2=0\\(1-2y)(1+2y)=0$

⇒  $1-2y=0$   or   $1+2y=0$

⇒ $2y=1$  or   $2y=-1$

⇒ $y= \frac{1}{2}$  or  $y=\frac{-1}{2}$

P/s:   $A^2-B^2=(A-B)(A+B)$

Ok done. Thank to me :>

Answer 2

Answer: y = 1/2 or y = -1/2

Step-by-step explanation:

First move the 1 to the side with the 0

-4y^2 = -1 // -1 because when you move a number to the other side of the equation  you have to multiply -1

Then divide by -4 on each side

y^2 = 1/4

square root each side

|y| = 1/2 // sq rt y^2 becomes an absolute value because when you square root a square it automatically becomes an absolute value

so the answer would be y = 1/2 or -1/2

to check just put it into the equation as y. Both work.