Question

Which of the following are solutions to the equation 3x2 + 7x + 4 = 0
Select all that apply
O x=-1
O x=-4/3
x=3/4
x=1

Answer 1

For this case we have the following quadratic equation:

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

Where:

$a = 3\\b = 7\\c = 4$

According to the quadratic formula we have:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Substituting:

$x = \frac {-7 \pm \sqrt {7 ^ 2-4 (3) (4)}} {2 (3)}\\x = \frac {-7 \pm \sqrt {49-48}} {6}\\x = \frac {-7 \pm \sqrt {1}} {6}\\x = \frac {-7 \pm1} {6}$

We have two roots:

$x_ {1} = \frac {-7 + 1} {6} = \frac {-6} {6} = - 1\\x_ {2} = \frac {-7-1} {6} = \frac {-8} {6} = - \frac {4} {3}$

Answer:

$x_ {1} = - 1\\x_ {2} = - \frac {4} {3}$

Answer 2

Answer:

x = -4/3            x=-1

Step-by-step explanation:

3x^2 + 7x + 4 = 0

Factor the equation

(3x+4) (x+1) = 0

Using the zero product property

3x+4 =0         x+1 =0

3x+4-4=0-4    x+1-1=0-1

3x=-4                     x=-1

3x/3 = -4/3

x = -4/3            x=-1