How do you solve the quadratic equation by completing the square: #y^2 + 16y = 2#?

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Answer 1 · 2018-06-14T17:57:09

#y= -8+-sqrt(66)#

to complete the square you use the formula:

#ax^2+bx+c#

a must equal 1

#c=(b/2)^2#

the completed square is:

#(x+b/2)^2#

Here we go, in your function the y is the general formula's x:

#y^2 + 16y = 2#

#y^2 + 16y +underbrace(c = 2+c)#
we add c to both sides so we don't alter the equation

now solve c:

#c=(b/2)^2 = (16/2)^2=64#

#y^2 + 16y +64 = 2+64#

now complete the square:

#(y+8)^2 = 66#

Now solve:

#sqrt((y+8)^2) = +-sqrt(66)#

#y+8 = +-sqrt(66)#

#y= -8+-sqrt(66)#