Answer :
In this type of equation you could use different methods. But I choose to do Quadratic Method.
a= 3, b= 8, and c= 4.
X= -(8)+_SQUARE ROOT (8)^2-4(3)(4)/2(3)
X= -8+_SQUARE ROOT16/6 Square Root of 16 is 4
X= -8+_4/6
X= -8+4/6= -4/6= -2/3.
X= -8-4/6= -12/6= -2.
Answer is: -2/3 and -2.
3x² + 8x + 4
First, i divide the equation into two parenthesis, so that the first parts of both multiply to make the first term, 3x². 3x * x = 3x²
(3x + )(x + )
Then I find two numbers that multiply to make 4, which are 1 and 4, or 2 and 2.
Our options are:
(3x + 1)(x + 4)
(3x + 4)(x + 1)
(3x + 2)(3x +2)
To figure out which one to use, I'm just going to FOIL them all.
(3x + 1)(x + 4) = 3x² + 1x + 12x + 4 = 3x² + 13x + 4
(3x + 4)(x + 1) = 3x² + 4x + 3x + 4 = 3x² + 7x + 4
(3x + 2)(x +2) = 3x² + 2x + 6x + 4 = 3x² + 8x + 4
The factored form is:
(3x + 2)(x + 2)
To solve for the roots ( where the graph crosses the x axis, where y = 0) we set the equation equal to 0:
(3x + 2)(x + 2) = 0
The zero product property says that anything times 0 is 0, so we set each individual part equal to 0 and solve for the two roots.
3x + 2 = 0
3x = -2
x = -2/3
x + 2 = 0
x = -2
First, i divide the equation into two parenthesis, so that the first parts of both multiply to make the first term, 3x². 3x * x = 3x²
(3x + )(x + )
Then I find two numbers that multiply to make 4, which are 1 and 4, or 2 and 2.
Our options are:
(3x + 1)(x + 4)
(3x + 4)(x + 1)
(3x + 2)(3x +2)
To figure out which one to use, I'm just going to FOIL them all.
(3x + 1)(x + 4) = 3x² + 1x + 12x + 4 = 3x² + 13x + 4
(3x + 4)(x + 1) = 3x² + 4x + 3x + 4 = 3x² + 7x + 4
(3x + 2)(x +2) = 3x² + 2x + 6x + 4 = 3x² + 8x + 4
The factored form is:
(3x + 2)(x + 2)
To solve for the roots ( where the graph crosses the x axis, where y = 0) we set the equation equal to 0:
(3x + 2)(x + 2) = 0
The zero product property says that anything times 0 is 0, so we set each individual part equal to 0 and solve for the two roots.
3x + 2 = 0
3x = -2
x = -2/3
x + 2 = 0
x = -2
