How do you solve $4 y^{2} = 25$ $4y^2 = 25$ ?

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Answer 1 · 2016-03-29T16:26:13

The solutions are:

$y = - \frac{5}{2}$ $color(blue)(y=-5/2$

$y = \frac{5}{2}$ $color(blue)(y=5/2$

$4 y^{2} = 25$ $4y^2 = 25$

$4 y^{2} - 25 = 0$ $4y^2 - 25 = 0$

${(2 y)}^{2} - 5^{2} = 0$ $(2y)^2 - 5^2 = 0$

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

${(2 y)}^{2} - 5^{2} = (2 y + 5) (2 y - 5)$ $(2y)^2 - 5^2 = color(blue)((2y+5)(2y-5)$

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

$2 y + 5 = 0, y = - \frac{5}{2}$ $2y + 5 = 0, color(blue)(y=-5/2$

$2 y - 5 = 0, y = \frac{5}{2}$ $2y- 5 = 0, color(blue)(y=5/2$

How do you solve 4y^2 = 254y2=254y^2 = 25?