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

Question

Abigail Allen · Accepted Answer

#x = (5 +- i*sqrt(2))/3# First, square #(x-2)# and combine like terms. #y = 3(x-2)^2 + 2x - 3# #y = 3(x-2)(x-2) + 2x - 3# #y = 3(x^2 - 4x + 4) + 2x - 3# #y = 3x^2 - 12x + 12 + 2x - 3# #y = 3x^2 - 10x + 9# Quadratic Formula: #x = (-b +- sqrt(b^2 - 4ac))/(2a)# #x = (-(-10) +- sqrt((-10)^2 - 4(3)(9)))/(2(3))# #x = (10 +- sqrt(100 - 108))/6# #x = (10 +- sqrt(-8))/6# #x = (10 +- 2i*sqrt(2))/6# Simplify: #x = (5 +- i*sqrt(2))/3#

Julia Adams · Answer

undefined To find the roots of the quadratic equation $y = 3(x - 2)^2 + 2x + 3$, we first rewrite it in the standard form: $ax^2 + bx + c = 0$. Then we can identify $a$, $b$, and $c$ to apply the quadratic formula:

$y = 3(x - 2)^2 + 2x + 3$

$= 3(x^2 - 4x + 4) + 2x + 3$

$= 3x^2 - 12x + 12 + 2x + 3$

$= 3x^2 - 10x + 15$

Now, $a = 3$, $b = -10$, and $c = 15$.

Using the quadratic formula, $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$,

we substitute the values of $a$, $b$, and $c$ into the formula:

$x = \frac{{-(-10) \pm \sqrt{{(-10)^2 - 4 \cdot 3 \cdot 15}}}}{{2 \cdot 3}}$

Solving under the square root:  
$b^2 - 4ac = (-10)^2 - 4 \cdot 3 \cdot 15 = 100 - 180 = -80$

Since $b^2 - 4ac = -80$ is negative, the roots are imaginary.

Thus, the roots of the equation $y = 3(x - 2)^2 + 2x + 3$ are complex numbers, given by:

$x = \frac{{10 \pm i\sqrt{80}}}{{6}}$