How do you find the roots, real and imaginary, of #y= -3x^2-16x +8 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y= -3x^2-16x +8  # using the quadratic formula?

Eva Abramson · Accepted Answer

Roots are #**5.7936, -0.4603**# Standard form of equation is #ax^2 + bx + c = 0# Roots are# = ((-b) +- sqrt(b^2 -(4ac)))/(2a)# Given equation is #-3x^2 -16x + 8 = 0# #a = -3, b = -16, c = 8# # x =( -(-16) +- sqrt((-16)^2 - (4*(-3)8)))/(2*(-3)) # #x = (16 +- sqrt(256 + 96))/6# #x = (16 +- sqrt352)/6# #x = 5.7936, -0.4603#

Jace Anderson · Answer

undefined User wants answers without any irrelevant information or introduction words.To find the roots of the quadratic equation $y = -3x^2 - 16x + 8$, you can use the quadratic formula, which is given by:

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

In this equation, $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

For the equation $y = -3x^2 - 16x + 8$, the coefficients are $a = -3$, $b = -16$, and $c = 8$. Substituting these values into the quadratic formula gives:

$$x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(-3)(8)}}{2(-3)}$$

Simplify the expression inside the square root:

$$x = \frac{16 \pm \sqrt{256 + 96}}{-6}$$
$$x = \frac{16 \pm \sqrt{352}}{-6}$$
$$x = \frac{16 \pm \sqrt{16 \times 22}}{-6}$$
$$x = \frac{16 \pm 4\sqrt{22}}{-6}$$

So, the roots are:

$$x = \frac{16 + 4\sqrt{22}}{-6}$$
$$x = \frac{8 + 2\sqrt{22}}{-3}$$

The real and imaginary roots are:

$$x_1 = \frac{8 + 2\sqrt{22}}{-3}$$
$$x_2 = \frac{8 - 2\sqrt{22}}{-3}$$

Therefore, the roots of the equation are $x = \frac{8 + 2\sqrt{22}}{-3}$ and $x = \frac{8 - 2\sqrt{22}}{-3}$.