★ QUADRATIC RESOLUTION ★
Solve : x^2 + x - (a+1)(a+2) =0
★ CONTENT QUALITY SUPPORT REQUIRED ★
Answers
Answered by
8
Hey there!
We have ,
x^2 + x - (a+1)(a+2) = 0
x^2 + x {(a+2 ) -(a+1) } - (a+1)(a+2) =0
{ x^2 + x (a+2)} -x (a +1) -(a + 1) (a+2)=0
x { x +(a+2) } - (a+1) { x+ (a+2)} =0
{x + (a+2) } { x - (a+1)} =0
x + (a +2) = 0 or , x - (a+1) =0
So, x = -(a+2) or, x = (a+1)
We have ,
x^2 + x - (a+1)(a+2) = 0
x^2 + x {(a+2 ) -(a+1) } - (a+1)(a+2) =0
{ x^2 + x (a+2)} -x (a +1) -(a + 1) (a+2)=0
x { x +(a+2) } - (a+1) { x+ (a+2)} =0
{x + (a+2) } { x - (a+1)} =0
x + (a +2) = 0 or , x - (a+1) =0
So, x = -(a+2) or, x = (a+1)
Anonymous:
★ CONTENT QUALITY SUPPORT ★
★ PERFECTION ★
I think..a simpler solution should have been by checking sum and product of roots
As product got -ve sign we will check the sum of roots to assign -ve sign to one of them.
Answered by
4
x^2 + x + (a+1)(a+2) = 0
Splitting the middle term (x)
x^2 + [(a+2) - (a+1)]x + (a+1)(a+2) = 0
x^2 + (a+2)x - (a+1)x + (a+1)(a+2) = 0
x(x+(a+2) - (a+1)(x + (a+2)) = 0
(x-(a+1)) (x+(a+2)) = 0
x-(a+1) = 0 x+(a+2)) = 0
x = a+1 x = -(a+2)
= -a - 2
Splitting the middle term (x)
x^2 + [(a+2) - (a+1)]x + (a+1)(a+2) = 0
x^2 + (a+2)x - (a+1)x + (a+1)(a+2) = 0
x(x+(a+2) - (a+1)(x + (a+2)) = 0
(x-(a+1)) (x+(a+2)) = 0
x-(a+1) = 0 x+(a+2)) = 0
x = a+1 x = -(a+2)
= -a - 2
Similar questions