How do you find the roots, real and imaginary, of $y=(x +4)^2+4x^2-3x + 2$ using the quadratic formula?

Question

1518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-03T17:28:02

See a solution process below:

First, we need to write the right side of the equation in standard form:
$y = (x + 4)^2 + 4x^2 - 3x + 2$

$y = x^2 + 8x + 16 + 4x^2 - 3x + 2$

$y = 1x^2 + 4x^2 + 8x - 3x + 16 + 2$

$y = (1 + 4)x^2 + (8 - 3)x + (16 + 2)$

$y = 5x^2 + 5x + 18$

We can now use the quadratic equation to solve this problem:

For $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$color(red)(5)$ for $color(red)(a)$

$color(blue)(5)$ for $color(blue)(b)$

$color(green)(18)$ for $color(green)(c)$ gives:

$x = (-color(blue)(5) +- sqrt(color(blue)(5)^2 - (4 * color(red)(5) * color(green)(18))))/(2 * color(red)(5))$

$x = (-color(blue)(5) +- sqrt(25 - 360))/10$

$x = (-5 +- sqrt(-335))/10$

How do you find the roots, real and imaginary, of y=(x +4)^2+4x^2-3x + 2y=(x +4)^2+4x^2-3x + 2 using the quadratic formula?