How do you solve #x^2 - 3x = 18# by completing the square?

2 Answers
Jul 14, 2018

Answer:

#x=-3" or "x=6#

Explanation:

#"add "(1/2"coefficient of the x-term")^2" to both sides"#

#x^2+2(-3/2)x color(red)(+9/4)=18color(red)(+9/4)#

#(x-3/2)^2=81/4#

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

#sqrt((x-3/2)^2)=+-sqrt(81/4)larrcolor(blue)"note plus or minus"#

#x-3/2=+-9/2#

#"add "3/2" to both sides"#

#x=3/2+-9/2#

#x=3/2-9/2=-3" or "x=3/2+9/2=6#

Jul 14, 2018

Answer:

#x=6 or x=-3#

Explanation:

Here ,

#x^2-3x=18#

#=>x^2-3x+k=k+18...to(I)#

We have to find #k# such that ,#x^2-3x+k#
forms a perfect square ,where

#color(blue)(1^(st)term=x^2# ,

#color(blue)(2^(nd)term=-3x#

#color(blue)(3^(rd)term=k#

Formula to find #3^(rd)term# is :

#color(red)(3^(rd)term=(2^(nd)term)^2/(4 xx1^(st)term))...to(A)#

So,

#k=(-3x)^2/(4*x^2)=(9x^2)/(4x^2)=9/4#

Subst. #k=9/4 # into #(I)# ,we get

#=>x^2-3x+9/4=9/4+18=81/4#

#=>(x-3/2)^2=(9/2)^2#

#=>x-3/2=+-9/2#

#=>x-3/2=9/2 orx-3/2=-9/2#

#=>x=9/2+3/2 or x=3/2-9/2#

#=>x=12/2 or x=-6/2#

#=>x=6 or x=-3#

......................................................................................

Note:
We can use Formula #(A)# without any doubt.