How do you solve #\sqrt { - 80+ 18r } = r#?

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Answer 1 · 2017-03-04T01:11:56

#r=sqrt161-9#
#r=-sqrt161-9#

Given

#sqrt(-80+18r)=r#

Taking square on both sides we get -

#-80+18r=r^2#

By rearranging we get -

#r^2+18r-80=0#

Use completing the square method to solve for #r#

#r^2+18r=80#

#r^2+18r+81=80+81#

#(r+9)^2=161#

#r+9=+-sqrt(161)#

#r=sqrt161-9#
#r=-sqrt161-9#