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

Question

Kennedy Adams · Accepted Answer

See a solution process below:

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#

Hazel Anglin · Answer

undefined To find the roots of the quadratic equation $ y = 5x^2 - x - (x - 1)^2 $, we first rewrite it in the standard form $ ax^2 + bx + c = 0 $. Then, we apply the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Where $ a = 5 $, $ b = -1 $, and $ c = -(x - 1)^2 $.

Substituting these values into the quadratic formula:

$$ x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4(5)(-(x - 1)^2)}}}}{{2(5)}} $$

$$ x = \frac{{1 \pm \sqrt{{1 + 20(x - 1)^2}}}}{{10}} $$

The roots will be real if the discriminant $ b^2 - 4ac $ is non-negative, and imaginary if it's negative.