How do you solve #2x + sqrt(x+1) = 8#?

1 Answer
Prof_S
Mar 2, 2016

#x=3# or #x=5.25#

Explanation:

Subtract #2x# from bopth sides of the equation, yielding #sqrt(x+1) = 8 -2x#. Square both sides of the equation, yielding #x+1=64-32x+4x^2#. Subtract x+1 from both sides of the equation, yielding # 0=63-33x+4x^2#.

Using the quadratic equation, one gets
# x = (33 (+/-) sqrt(1089 - 4*4*63))/8 = (33 (+/-) sqrt(81))/8 = (33 (+/-) 9)/8#.
So, #x = 42/8# or #x=24/8#.

Related questions
See all questions in Radical Equations
Impact of this question
5153 views around the world
You can reuse this answer
Creative Commons License