How do you solve #2y ^ { 2} = y + 36#?

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Answer 1 · 2017-06-01T01:57:28

See a solution process below:

First, subtract #color(red)(y)# and #color(blue)(36)# from each side of the equation to form a standard quadratic equation:

#2y^2 - color(red)(y) - color(blue)(36) = y - color(red)(y) + 36 - color(blue)(36)#

#2y^2 - y - 36 = 0 + 0#

#2y^2 - y - 36 = 0#

Next, factor the quadratic as:

#(2y - 9)(y + 4) = 0#

Now solve each term on the left for #0#:

Solution 1)

#2y - 9 = 0#

#2y - 9 + color(red)(9) = 0 + color(red)(9)#

#2y - 0 = 9#

#2y = 9#

#(2y)/color(red)(2) = 9/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = 9/2#

#y = 9/2#

Solution)

#y + 4 = 0#

#y + 4 - color(red)(4) = 0 - color(red)(4)#

#y + 0 = -4#

#y = -4#

The solution: #y = 9/2#; #y = -4#