How do you factor #4- x ^ { 2} + 16y ^ { 2} = 0#?

1 Answer
Feb 4, 2018

See a solution process below:

Explanation:

First, subtract #color(red)(4)# and add #color(blue)(x^2)# to each side of the equation to isolate the #y# term while keeping the equation balanced:

#4 - color(red)(4) - x^2 + color(blue)(x^2) + 16y^2 = 0 + color(blue)(x^2) - color(red)(4)#

#0 - 0 + 16y^2 = x^2 - 4#

#16y^2 = x^2 - 4#

Next, use this rule of quadratics to factor the right side of the equation:

#(color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y)) = color(red)(x)^2 - color(blue)(y)^2#

#16y^2 = color(red)(x)^2 - color(blue)(4)#

#16y^2 = color(red)(x)^2 - color(blue)(2)^2#

#16y^2 = (color(red)(x) + color(blue)(2))(color(red)(x) - color(blue)(2))#

If you are required to continue and solve for #y#, take the square root of each side of the equation giving:

#sqrt(16y^2) = +-sqrt((color(red)(x) + color(blue)(2))(color(red)(x) - color(blue)(2)))#

#4y = +-sqrt((color(red)(x) + color(blue)(2))(color(red)(x) - color(blue)(2)))#

Then, divide each side of the equation by #color(green)(4)# to solve for #y#:

#(4y)/color(green)(4) = +-sqrt((color(red)(x) + color(blue)(2))(color(red)(x) - color(blue)(2)))/color(green)(4)#

#y = +-sqrt((color(red)(x) + color(blue)(2))(color(red)(x) - color(blue)(2)))/color(green)(4)#