How do you find the zeros, real and imaginary, of #y= 3x^2+4x+2 # using the quadratic formula?

Question

Hazel Anglin · Accepted Answer

Substitute the coefficients into the equation and evaluate.

The quadratic equation is #x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}# In the example a = 3, b = 4 and c = 2 #x = (-4 +- sqrt(4^2 - 4*3*2))/(2*3)# #x= (-4 +- sqrt(16 - 24))/6# #x = -4/6 +- sqrt(-8)/6# #x = -2/3(1 +sqrt(-2))# or #-2/3(1 - sqrt(-2))# The equation only has imaginary roots. The graph never touches or crosses the x axis.

Hazel Anglin · Answer

undefined To find the zeros of the quadratic equation $y = 3x^2 + 4x + 2$, use the quadratic formula:

$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

Where $a = 3$, $b = 4$, and $c = 2$.

Plugging these values into the quadratic formula:

$$x = \frac{{-4 \pm \sqrt{{4^2 - 4 \cdot 3 \cdot 2}}}}{{2 \cdot 3}}$$

Simplify under the square root:

$$x = \frac{{-4 \pm \sqrt{{16 - 24}}}}{{6}}$$
$$x = \frac{{-4 \pm \sqrt{{-8}}}}{{6}}$$

Since the square root of a negative number results in an imaginary number, the zeros of the quadratic equation are imaginary.

$$x = \frac{{-4 \pm \sqrt{{-8}}}}{{6}}$$
$$x = \frac{{-4 \pm \sqrt{{8i}}}}{{6}}$$

Therefore, the zeros of the quadratic equation are complex numbers, where $x = \frac{{-4 + \sqrt{{8i}}}}{{6}}$ and $x = \frac{{-4 - \sqrt{{8i}}}}{{6}}$.