How do you solve r^2+11r+18=0 by completing the square?

2 Answers
Shwetank Mauria
Apr 29, 2018

#r=-2# or #r=-9#

Explanation:

As #(a+b)^2=a^2+2ab+b^2#, to complete a trinomial as a square of binomial,

we need two complete squares #a^2# and #b^2# and a middle term which here is #2ab# i.e. twice the product of square roots of these squares i.e. #2xxaxxb=2ab#.

Here in #r^2+11r+18=0#, we have #r^2# and then #11r=2xxrxx11/2#, hence we should have #(11/2)^2#, but we have #18#. So let us write #r^2+11r+18=0# as

#r^2+11r+(11/2)^2-(11/2)^2+18=0#, which can be written as

#[r^2+2xxrxx11/2+(11/2)^2]-121/4+18=0#

or #(r+11/2)^2-(121-72)/4=0#

or #(r+11/2)^2-49/4=0#

i.e. #(r+11/2)^2-(7/2)^2=0#

As it is in the form #a^2-b^2#, we can factorize it as #(a-b)(a+b)# i.e.

#(r+11/2-7/2)(r+11/2+7/2)=0#

or #(r+4/2)(r+18/2)-0# i.e. #(r+2)(r+9)=0#

Hence either #r+2=0# i.e. #r=-2# or #r+9=0# i.e. #r=-9#.

Jim G.
Apr 29, 2018

#r=-9" or "r=-2#

Explanation:

#"to solve using the method of "color(blue)"completing the square"#

#• " the coefficient of the "r^2" term must be 1 which it is"#

#• " add/subtract "(1/2"coefficient of the r-term")^2" to"#
#r^2+11r#

#rArrr^2+2(11/2)r color(red)(+121/4)color(red)(-121/4)+18=0#

#rArr(r+11/2)^2-49/4=0#

#rArr(r+11/2)^2=49/4#

#color(blue)"take the square root of both sides"#

#sqrt((r+11/2)^2)=+-sqrt(49/4)larrcolor(blue)"note plus or minus"#

#rArrr+11/2=+-7/2#

#rArrr=-11/2+-7/2#

#rArrr=-11/2-7/2=-9" or "r=-11/2+7/2=-2#

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