How do you solve this: x2 + 5x + 4 = 0?

Let's use the Completing the Square method. It's more convenient.

x² + 5x + 4 = 0

x² + 5x = - 4

Divide the second term of the binomial (5x) by 2x and square the quotient to get the third term.

x² + 5x + (5x / 2x)² = - 4

x² + 5x + 6.25 = - 4 (Transfer 6.25)

x² + 5x + 6.25 = - 4 + 6.25 (Factor trinomial)

(x + 2.5)² = 2.25 (Square root)

x + 2.5 = ± 1.5 (Transpose 2.5)

x = (-2.5 + 1.5) , (-2.5 - 1.5)

x = - 1, - 4 (Answers)

Hope this helps.

How do you factorise x2+5x+4 (x+4)(x+1) And solve x2+5x+4=0? (x+4)(x+1)=0 (x+4)=0 x=-4 (x+1)=0 x=-1

You need to factor x. To do that, find 2 numbers that both add up to 5 and multiply to 4. That should be easy, it's just 1 and 4! So you have factored it to (x+4)(x+1)=0. Now we know that 2 things multiplied together to make 0 can only happen when either the first thing is 0 or the second thing is 0, so we have two solutions:

x+4=0

x=-4

or

x+1=0

x=-1

ok so there are 2 ways --- firstly by just sorta looking at it --- you have to get 2 numbers ... (both multiples of 4) that when added or subtracted = 5 ... so uve got: ... 1,4 and 2,2 ... it obviously isnt 2,2 ... so uve got 1,4 ... because 1+4 = 5 ... so now u put them in 2 brackets ... (x+1)(x+4) and ur solutions are therefore x= -1 and x=- 4 ... or alternately use the quadratic formula ... for an equation in the for ax^2 +bx +c the answer is

[-b + or - sqrt(b^2 -4*a*c)]/2a ... so in this instance

(-5 + sqrt (25 - 16) )/2... 25 - 16 = 9 ... sqrt 9 = 3 so -5+3 = -2/2 = -1

and (-5 - sqrt(25 - 16)/2 = -5 -3 /2 = -8/2 = -4

x^2+5x+4=0

find factors of 4 which will produce a positive 5 x

(x+1)(x+4)=0

x=-1 or -4

x^2 + 5x + 4 = 0

x^2 + 4x + x + 4 = 0

(x^2 + 4x) + (x + 4) = 0

x(x + 4) + 1(x + 4) = 0

(x + 4)(x + 1) = 0

x + 4 = 0

x = -4

x + 1 = 0

x = -1

∴ x = -4, -1

you can simply solve it by scientific calculator

if you don't have, trial and error can help