Question

How do you complete the square to solve `x^2+x=19/4`?

Answer 1

`x=-1/2+sqrt5`
`x=-1/2-sqrt5`

Explanation:

Given -

`x^2+x=19/4`

`x^2+x+1/4=19/4+1/4=(19+1)/4=20/4`

`(x+1/2)^2= 20/4`

`(x+1/2)=+-sqrt (20/4)=+-(sqrt5xxcancelsqrt(4))/cancelsqrt(4)`
`x+1/2=+-sqrt 5`

`x=-1/2+sqrt5`
`x=-1/2-sqrt5`

Answer 2

`x=-1/2-sqrt5` or `x=-1/2+sqrt5`

Explanation:

Remember the identity `a^2+2xxaxxb+b^2=(a+b)^2`. We have

`x^2+x=19/4`

or `x^2+2xx x xx 1/2=19/4`

Hence to complete square we must add `(1/2)^2` on both sides, as comparing it with above identity, we get `b=1/2` and our equation becomes

`x^2+2xx x xx 1/2+(1/2)^2=19/4+1/4`

or `(x+1/2)^2=5=(sqrt5)^2`

or `(x+1/2)^2-(sqrt5)^2=0`

and using the identity `a^2-b^2=(a+b)(a-b)`, this becomes

`(x+1/2+sqrt5)(x+1/2-sqrt5)=0`

Hence either `x+1/2+sqrt5=0` i.e. `x=-1/2-sqrt5`

or `x+1/2-sqrt5=0` i.e. `x=-1/2+sqrt5`