How do you find the roots, real and imaginary, of #y= 12x^2+3x-6 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y= 12x^2+3x-6  # using the quadratic formula?

Avery Alexander · Accepted Answer

#x = -1/8+- sqrt33/8# #y = 12x^2 + 3x - 6 = 4x^2 + x - 2 = 0# #D = d^2 = b^2 - 4ac = 1 + 32 = 33# --> # d = +-sqrt33# There are 2 real roots: #x = -b/(2a) +- d/(2a) = -1/8 +- sqrt33/8#

Julia Adams · Answer

undefined To find the roots of the quadratic equation $y = 12x^2 + 3x - 6$ using the quadratic formula, $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = 12$, $b = 3$, and $c = -6$:

1. Substitute the values of $a$, $b$, and $c$ into the quadratic formula.
2. Calculate the discriminant, $b^2 - 4ac$.
3. Determine whether the discriminant is positive, negative, or zero.
4. If the discriminant is positive, there are two real roots. Use the formula to find both roots.
5. If the discriminant is zero, there is one real root (a repeated root).
6. If the discriminant is negative, there are two complex roots (imaginary roots). Use the formula to find both roots.

Substituting the values:
$a = 12$, $b = 3$, $c = -6$

$$\text{Discriminant} = b^2 - 4ac$$
$$\text{Discriminant} = (3)^2 - 4(12)(-6)$$
$$\text{Discriminant} = 9 + 288$$
$$\text{Discriminant} = 297$$

Since the discriminant is positive, there are two real roots.

Using the quadratic formula:
$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$
$$x = \frac{{-3 \pm \sqrt{297}}}{{2(12)}}$$

The roots are:

$$x_1 = \frac{{-3 + \sqrt{297}}}{{24}}$$
$$x_2 = \frac{{-3 - \sqrt{297}}}{{24}}$$