Question

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

Answer 1

`y=-1 or y=-2`

Explanation:

To solve this equation first,we should factorize it.
`" "`
In this exercise we can't factorize using common factor nor applying
`" "`
Polynomial identities , so I will apply trial and error method instead.
`" "`
First: Factorize
`" "`
Trial and error method :
`" "`
`X^2+SX+P=0`
`" "`
if we can find two real numbers `a and b` whose sum is S and
`" "`
product is P then `color(blue)(X^2+SX+P=0=(X+a)(X+b)`
`" "`
In the given equation the polynomial
`" "`
`y^2+3y+2` has
`" "`
`S=3 and P=2`
`" "`
So,
`" "`
`color(blue)(y^2+3y+2=(y+2)(y+1))`

second: solve

`y^2+3y+2=0`
`" "`
`(y+2)(y+1)=0`
`" "`
`y+2=0rArry=-2`
`" "`
Or
`" "`
`y+1=0rArry=-1`