How do you find the roots, real and imaginary, of #y=-2(x +3)^2-5x + 6# using the quadratic formula?

Question

Hazel Anglin · Accepted Answer

Roots: #x in {-7.723, -0.777}# First convert the given equation: #y=-2(x+3)^2-5x+6# into standard quadratic form: #y=-2(x^2+6x+9)-5x+6# #y=-2x^2-12x-18-5x+6# #y=-2x^2-17x-12# Apply the quadratic formula: #color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)# with #a=-2#, #b=-17#, and #c=-12# #x=(17+-sqrt(17^2-4(2)(12)))/(2(-2)# which (with a bit of arithmetic) gives: #color(white)("XXX")x~~-7.723# or #x~~-0.777#

Eva Abramson · Answer

undefined To find the roots of the quadratic equation $y = -2(x + 3)^2 - 5x + 6$, we first need to rewrite it in the standard form $ax^2 + bx + c = 0$. After that, we can use the quadratic formula, which states that for an equation of the form $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In this case, we have $y = -2(x + 3)^2 - 5x + 6$. Expanding and simplifying, we get $y = -2(x^2 + 6x + 9) - 5x + 6$, which further simplifies to $y = -2x^2 - 12x - 18 - 5x + 6$, or $y = -2x^2 - 17x - 12$.

Comparing this to the standard form $ax^2 + bx + c = 0$, we have $a = -2$, $b = -17$, and $c = -12$.

Plugging these values into the quadratic formula, we get:

$$x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(-2)(-12)}}{2(-2)}$$

Simplifying further:

$$x = \frac{17 \pm \sqrt{289 - 96}}{-4}$$

$$x = \frac{17 \pm \sqrt{193}}{-4}$$

So, the roots are:

$$x = \frac{17 + \sqrt{193}}{-4}$$

$$x = \frac{17 - \sqrt{193}}{-4}$$

These are the real and imaginary roots of the given quadratic equation.