Question

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

Answer 1

`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`

Answer 2

`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.