Solve x (squared) +5x+4=0
Posted Sun 29th May, 2011 @ 18:30 by
summer2011
This is how I do it, very clear steps so it should be easy to understand, (x^2 is x squared):
x^2+5x+4 = 0
x^2+x+4x+4 = 0
x(x+1)+4(x+1) = 0
(x+4) (x+1) = 0
x=-4 or x=-1
Answered Sun 29th May, 2011 @ 22:12 by
N. M
@Falak Hussain
" So the first thing you have to do is factor
x (square) + 5x + 6 = (x + 4)(x + 1)
Set this equal to zero
(x + 4)(x + 1) = 0
Solve each factor
x + 4 = 0 x + 1 = 0
x = 4 x = 1 "
WRONG
x+4=0 and x+1=0
Therefore x=-4 and x=-1
Answered Sun 29th May, 2011 @ 20:44 by
ChrisEdited by
Chris on Sun 29th May, 2011 @ 20:50
Okay so you have to do factor and sum, which two numbers multiply to get 4 , and add to get 5?
The answer is 4 and 1 , so its (x+4)(x+1) ;)
Answered Sun 29th May, 2011 @ 18:35 by
Lamise Hassan
This is a classic example of a Quadratic Equation
x² + 5x + 4 = 0
1. Look at the constant (+4 in this case). Find its factors:
1x4=4
2x2=4
therefore we might say (x+2)(x+2)=0
This is incorrect since when expanded:
*** + Xx2 + Xx2 + 4 = X² + 4x + 4
Therefore the quadratic must include the OTHER paired factors (1 and 4)
(x+1)(x+4)=0
To solve,
(x+1) = 0 or (x+4)=0
x=-1 or x=-4
Answered Sun 29th May, 2011 @ 20:49 by
ChrisEdited by
Chris on Sun 29th May, 2011 @ 20:51
factorise it. then make each bracket =0
Answered Mon 30th May, 2011 @ 11:03 by
Georgina