How do you solve #\sqrt { 2z + 33} + 2= z + 1#?

1 Answer
iceman
Sep 17, 2017

#z=8#

Explanation:

#sqrt(2z+33)+2=z+1# => subtract 2 from both sides:
#sqrt(2z+33)=z-1# => square both sides:
#2z+33=z^2-2z+1# => subtract 2z and 33 from both sides:
#z^2-4z-32=0# => factor:
#(z+4)(z-8)=0#
#z=-4, 8# => all roots must be checked in the original equation:

#z=-4:#
#sqrt(2*(-4)+33)+2=-4+1#
#sqrt(33-8)+2=-3#
#sqrt25+2=-3#
#5+2=-3#
#7=-3# => obviously 7 is not equal to -3, so this is always false, therefore #z=-4# is an Extraneous root and must be rejected

#z=8:#
#sqrt(2*(8)+33)+2=8+1#
#sqrt(16+33)+2=9#
#sqrt49+2=9#
#7+2=9#
#9=9# => verified #z=8# is a valid root thus:
#z=8# => the only valid solution

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