Question

find the roots of the equation x²+x-(m+2)(m+1)=0​

Answer 1

$\bf \color{brown}{ \underline{ \underline{Solution :}}} \\ \\ \\ \tt \: {x}^{2} + x - (m + 2)(m + 1) = 0 \\ \\ \implies \tt \: {x}^{2} + (m + 2 )x- (m + 1)x - (m + 2)(m + 1) = 0 \\ \\ \implies \tt \: x(x + m + 2) - (m + 1)(x + m + 2) = 0 \\ \\ \implies \tt (x + m + 2)(x - m - 1) = 0 \\ \\ \begin{array}{c|c} \implies \tt (x + m + 2) = 0& \implies \tt(x - m - 1) = 0 \\ \\ \\ \tt \therefore\green{ \underline{ \boxed{\tt{x = - (m + 2)}}}}& \therefore\green{ \underline{ \boxed{\tt{x = (m + 1)}}}}\end{array} \\ \\ \\ \\ \tt \colorbox{aqua}{@StayHigh}$

Answer 2

$\huge{\underline{\underbrace{\mathcal\color{gold}{Answer}}}}$

$\blue{x^2+x−(m+2)(m+1)=0}$

⏭$⟹\green{x^2+(m+2)x−(m+1)x−(m+2)(m+1)=0}$

⏭$⟹\purple {x~(x+m+2)−(m+1)~(x+m+2)~=0}$

⏭$⟹\red{(x+m+2)~(x−m−1)=0}$

$⟹(x+m+2)=0\\∴x=−(m+2)$

$⟹(x−m−1)=0\\∴x=(m+1)$