How do you solve #x^2+3x+2=0#?

2 Answers
Jul 8, 2015

Answer:

The solutions for the equation are:

#color(blue)(x=-1, x=-2#

Explanation:

#x^2 + 3x +2 =0#

We can solve the expression by first factorising.

Factorising by splitting the middle term

#x^2 + 3x +2 =0#

#x^2 + 2x + x+2 =0#

#x(x+ 2) +1 (x+2) =0#

#color(blue)((x+1)(x+ 2)=0#

Equating the factors with zero:
#color(blue)(x+1 = 0 , x =-1)#
#color(blue)(x+ 2=0 , x=-2#

Jul 8, 2015

Answer:

x=-2 or x=-1

Explanation:

Two standard ways to solve a quadratic equation:

Firstly you could factorise it to the form:-
#x^2+3x+2=0#
#x^2+(a+b)x+ab=0#
#(x+a)(x+b)=0#
Therefore we need two numbers that satisfy:-
#a+b=3 & ab=2#
#=> a=2; b=1#

So the expression is:-
#(x+2)(x+1)=0#
It's then trivial to see that if #x=-2 or x=-1# then the expression is true. These are the solutions.

The other solution is to use the formula for the solution of a quadratic equation:
#a*x^2+b*x+c=0#
=>
#x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=1, b=3, c=2# so we have:
#x=(-3+sqrt(9-8))/2=-1# or #x=(-3-sqrt(9-8))/2=-2#
The same two solutions