Solve for : x
\( x^2 + 4x + 3 = -(x + 3)(x + 5). \)
Thanks help when you can :)
Factor L side to (x+3)(x+1) then cancel (x+3) from both sides of the equation and solve.... x+1 = - (x+5)
Not sure why you say it is incorrect,
if you solve as instructed: x+1 = - (x+5) you will find x = -3
I think YOU are incorrect... thanx for not really trying though !
You can only cancel x+3 from borh sides (i.e. divide both sides by x+3) as long as x+3 is not zero. Here, x+3 turns out to be zero.
Fortuitously, you still get the right answer in this case! However, performing mathematically illegal operations is not valid, whatever the apparent result.
Solve for x:
x^2 + 4 x + 3 = -(x + 3) (x + 5)
Expand out terms of the right hand side:
x^2 + 4 x + 3 = -x^2 - 8 x - 15
Add x^2 + 8 x + 15 to both sides:
2 x^2 + 12 x + 18 = 0
Divide both sides by 2:
x^2 + 6 x + 9 = 0
Write the left hand side as a square:
(x + 3)^2 = 0
Take the square root of both sides:
x + 3 = 0
Subtract 3 from both sides:
x = -3