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

2 Answers
Jun 12, 2017

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

Jun 12, 2017

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