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

Jul 8, 2015

The solutions for the equation are:

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

#### Explanation:

${x}^{2} + 3 x + 2 = 0$

We can solve the expression by first factorising.

Factorising by splitting the middle term

${x}^{2} + 3 x + 2 = 0$

${x}^{2} + 2 x + x + 2 = 0$

$x \left(x + 2\right) + 1 \left(x + 2\right) = 0$

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

Equating the factors with zero:
$\textcolor{b l u e}{x + 1 = 0 , x = - 1}$
color(blue)(x+ 2=0 , x=-2

Jul 8, 2015

x=-2 or x=-1

#### Explanation:

Two standard ways to solve a quadratic equation:

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

So the expression is:-
$\left(x + 2\right) \left(x + 1\right) = 0$
It's then trivial to see that if $x = - 2 \mathmr{and} x = - 1$ then the expression is true. These are the solutions.

The other solution is to use the formula for the solution of a quadratic equation:
$a \cdot {x}^{2} + b \cdot x + c = 0$
=>
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 1 , b = 3 , c = 2$ so we have:
$x = \frac{- 3 + \sqrt{9 - 8}}{2} = - 1$ or $x = \frac{- 3 - \sqrt{9 - 8}}{2} = - 2$
The same two solutions