How do you find the roots, real and imaginary, of #y= 2x^2 - 2x -1 - (x-7)^2 # using the quadratic formula?

1 Answer
Nov 13, 2017

Answer:

See a solution process below:

Explanation:

First, we need to put the equation in standard form by:
1. Expanding the terms in parenthesis
2. Grouping like terms
3. Combining like terms

#y = 2x^2 - 2x - 1 - (x - 7)^2#

#y = 2x^2 - 2x - 1 - (x^2 - 7x - 7x + 49)#

#y = 2x^2 - 2x - 1 - x^2 + 7x + 7x - 49#

#y = 2x^2 - x^2 - 2x+ 7x + 7x - 1 - 49#

#y = 2x^2 - 1x^2 - 2x+ 7x + 7x - 1 - 49#

#y = (2 - 1)x^2 + (-2+ 7 + 7)x + (-1 - 49)#

#y = 1x^2 + 12x + (-50)#

#y = 1x^2 + 12x - 50#

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)(1)# for #color(red)(a)#

#color(blue)(12)# for #color(blue)(b)#

#color(green)(-50)# for #color(green)(c)# gives:

#x = (-color(blue)(12) +- sqrt(color(blue)(12)^2 - (4 * color(red)(1) * color(green)(-50))))/(2 * color(red)(1))#

#x = (-12 +- sqrt(144 - (-200)))/2#

#x = (-12 +- sqrt(144 + 200))/2#

#x = (-12 +- sqrt(344))/2#

#x = (-12 +- sqrt(4 * 86))/2#

#x = (-12 +- sqrt(4)sqrt(86))/2#

#x = (-12 +- 2sqrt(86))/2#

#x = -6 +- sqrt(86)#