Solved : `x^(2)+x-(a+2)(a+1)=0` by
(i) factorisation (ii) method of completing the square

Explanation

(i) By factorisation :
We have `x^(2)+x-(a+2)(a+1)=0`
`impliesx^(2)+"x"xx1-(a+2)(a+1)=0`
`impliesx^(2)+x[(a+2)-(a+1)]-(a+2)(a+1)=0`
`impliesx^(2)+x(a+2)-x(a+1)-(a+2)(a+1)=0`
`impliesx[x+(a+2)]-(a+1)[x+(a+2)]=0`
`implies[x+(a+2)][x-(a+1)]=0`
`:."eithre "x+(a+2)=0orx-(a+1)=0`
`impliesx=-(a+2)orx=(a+1)`
(ii) By the method of completing the square :
We have `x^(2)+x-(a+2)(a+1)=0`
`impliesx^(2)+x=(a+2)(a+1)`
Adding `((1)/(2))^(2)` on both side, we get
`x^(2)+x+((1)/(2))^(2)=a^(2)+3a+2+((1)/(2))^(2)`
`implies(x+(1)/(2))^(2)=(4a^(2)+12a+9)/(4)`
`implies(x=(1)/(2))^(2)=((2a+3)/(2))^(2)`
Taking square root of both sides, we get
`x+(1)/(2)=+-(2a+3)/(2)`
`:.x=(-1)/(2)+-(2a+3)/(2)=(-1+-(2a+3))/(2)`
`:.x=(-1+2a+3)/(2)orx=(-1-(2a+3))/(2)`
`impliesx=(2(a+1))/(2)orx=(-2(a+2))/(2)`
`impliesx=(a+1)orx=-(a+2)`