How do you find the roots, real and imaginary, of y= 5x^2-x-(x-1)^2 using the quadratic formula?

1 Answer
Nov 9, 2017

See a solution process below:

Explanation:

First, we need to expand the term in parenthesis on the right side of the equation using this rule:

(color(red)(x) - color(blue)(y))^2 = (color(red)(x) - color(blue)(y))(color(red)(x) - color(blue)(y)) = color(red)(x)^2 - 2color(red)(x)color(blue)(y) + color(blue)(y)^2

Substituting x for x and 1 for y gives:

y = 5x^2 - x - (x^2 - 2x + 1)

y = 5x^2 - x - x^2 + 2x - 1

We can next group and combine like terms:

y = 5x^2 - x^2 - x + 2x - 1

y = 5x^2 - 1x^2 - 1x + 2x - 1

y = (5 - 1)x^2 + (-1 + 2)x - 1

y = 4x^2 + 1x - 1

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

The quadratic formula states:

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)(4) for color(red)(a)

color(blue)(1) for color(blue)(b)

color(green)(-1) for color(green)(c) gives:

x = (-color(blue)(1) +- sqrt(color(blue)(1)^2 - (4 * color(red)(4) * color(green)(-1))))/(2 * color(red)(4))

x = (-color(blue)(1) +- sqrt(1 - (-16)))/8

x = (-color(blue)(1) +- sqrt(1 + 16))/8

x = (-color(blue)(1) +- sqrt(17))/8