# How do you solve #3x^2+7x+4=0#?

##### 1 Answer

See explanation for a few methods...

#### Explanation:

**Quick Method**

Notice that if you invert the sign of the term of odd degree then the sum of the coefficients is zero. That is:

We can deduce that

#0 = 3x^2+7x+4=(x+1)(3x+4)#

So the other root is

**AC Method**

Look for a pair of factors of

The pair

Use that pair to split the middle term and factor by grouping:

#0 = 3x^2+7x+4#

#=3x^2+3x+4x+4#

#=(3x^2+3x)+(4x+4)#

#=3x(x+1)+4(x+1)#

#=(3x+4)(x+1)#

Hence roots

**Completing the square**

Use the difference of squares identity too:

#a^2-b^2 = (a-b)(a+b)#

with

First multiply the equation by

#0 = 3x^2+7x+4#

becomes

#0 = 36x^2+84x+48#

#=(6x+7)^2-49+48#

#=(6x+7)^2-1^2#

#=((6x+7)-1)((6x+7)+1)#

#=(6x+6)(6x+8)#

#=(6(x+1))(2(3x+4))#

#=12(x+1)(3x+4)#

Hence roots