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

Answer:

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#