How do you find the zeros, real and imaginary, of #y=x^2 -34x+1# using the quadratic formula?

Question

1215 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-16T13:12:58

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)(-34)# for #color(blue)(b)#

#color(green)(1)# for #color(green)(c)# gives:

#x = (-color(blue)(-34) +- sqrt(color(blue)(-34)^2 - (4 * color(red)(1) * color(green)(1))))/(2 * color(red)(1))#

#x = (34 +- sqrt(1156 - 4))/2#

#x = (34 +- sqrt(1152))/2#

#x = (34 - sqrt(576 * 2))/2# and #x = (34 + sqrt(576 * 2))/2#

#x = (34 - sqrt(576)sqrt(2))/2# and #x = (34 + sqrt(576)sqrt(2))/2#

#x = (34 - 24sqrt(2))/2# and #x = (34 + 24sqrt(2))/2#

#x = 17 - 12sqrt(2)# and #x = 17 + 12sqrt(2)#