How do you find the zeros of # y = -x^2 +32x + 19 # using the quadratic formula?

Question

1442 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-29T20:22:47

See a solution process below:

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)(32)# for #color(blue)(b)#

#color(green)(19)# for #color(green)(c)# gives:

#x = (-color(blue)(32) +- sqrt(color(blue)(32)^2 - (4 * color(red)(-1) * color(green)(19))))/(2 * color(red)(-1))#

#x = (-color(blue)(32) +- sqrt(1024 - (-76)))/-2#

#x = (-color(blue)(32) +- sqrt(1024 + 76))/-2#

#x = (-color(blue)(32) +- sqrt(1100))/-2#

#x = (-color(blue)(32) - sqrt(100 * 11))/-2# and #x = (-color(blue)(32) + sqrt(100 * 11))/-2#

#x = (-color(blue)(32) - sqrt(100)sqrt(11))/-2# and #x = (-color(blue)(32) + sqrt(100)sqrt(11))/-2#

#x = (-color(blue)(32) - 10sqrt(11))/-2# and #x = (-color(blue)(32) + 10sqrt(11))/-2#

#x = (-color(blue)(32))/-2 - (10sqrt(11))/-2# and #x = (-color(blue)(32))/-2 + (10sqrt(11))/-2#

#x = 16 + 5sqrt(11)# and #x = 16 - 5sqrt(11)#