How do you find the roots, real and imaginary, of #y=-2(x+2)^2+16x^2-x-5 # using the quadratic formula?

1 Answer
Jul 24, 2017

Answer:

See a solution process below:

Explanation:

First, expand the term squared on the right side of the equation using the rule:

#(color(red)(a) + color(blue)(b))^2 = color(red)(a)^2 + 2color(red)(a)color(blue)(b) + color(blue)(b)^2#

Substituting #x# for #a# and #2# for #b# gives:

#y = -2(color(red)(x) + color(blue)(2))^2 + 16x^2 - x - 5#

#y = -2(color(red)(x)^2 + (2 * color(red)(x) * color(blue)(2)) + color(blue)(2)^2) + 16x^2 - x - 5#

#y = -2(x^2 + 4x + 4) + 16x^2 - x - 5#

Now, expand the term in parenthesis, group and combine like items:

#y = (-2 * x^2) + (2 * 4x) + (2 * 4) + 16x^2 - x - 5#

#y = -2x^2 + 8x + 8 + 16x^2 - x - 5#

#y = -2x^2 + 16x^2 + 8x - x + 8 - 5#

#y = (-2 + 16)x^2 + (8 - 1)x + (8 - 5)#

#y = 14x^2 + 7x + 3#

We can now use the quadratic equation to solve this problem. The quadratic formula states:

For #ax^2 + bx + 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)(14)# for #color(red)(a)#

#color(blue)(7)# for #color(blue)(b)#

#color(green)(3)# for #color(green)(c)# gives:

#x = (-color(blue)(7) +- sqrt(color(blue)(7)^2 - (4 * color(red)(14) * color(green)(3))))/(2 * color(red)(14))#

#x = (-color(blue)(7) +- sqrt(49 - 168))/28#

#x = (-color(blue)(7) +- sqrt(-119))/28#

Or

#x = -color(blue)(7)/28 +- sqrt(-119)/28#

#x = -1/4 +- sqrt(-119)/28#