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) + 5*x* + 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