Question

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

Answer 1

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`

Answer 2

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