How do you solve #y^2-8y-11 = x# for #y# in terms of #x# ?

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Answer 1 · 2015-08-09T21:05:08

Subtract #x# from both sides then use the quadratic formula to find:

#y = 4 +-sqrt(x+5)#

Subtract #x# from both sides to get:

#y^2 - 8y + (11-x) = 0#

Use the quadratic formula to solve for #y# in terms of #x#...

#y = (8+-sqrt(8^2-4(11-x)))/2#

#=(8+-sqrt(4(16-(11-x))))/2#

#=4+-sqrt(x+5)#