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#
