How do you find the zeros, real and imaginary, of #y=9x^2-8x-4 #using the quadratic formula?

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Answer 1 · 2017-08-21T01:20:22

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)(9)# for #color(red)(a)#

#color(blue)(-8)# for #color(blue)(b)#

#color(green)(-4)# for #color(green)(c)# gives:

#x = (-color(blue)((-8)) +- sqrt(color(blue)((-8))^2 - (4 * color(red)(9) * color(green)(-4))))/(2 * color(red)(9))#

#x = (8 +- sqrt(64 - (-144)))/18#

#x = (8 +- sqrt(64 + (144)))/18#

#x = (8 +- sqrt(208))/18#

#x = (8 +- sqrt(16 * 13))/18#

#x = (8 +- sqrt(16)sqrt(13))/18#

#x = (8 +- 4sqrt(13))/18#

#x = (8 - 4sqrt(13))/18# and #x = (8 + 4sqrt(13))/18#

#x = (4 - 2sqrt(13))/9# and #x = (4 + 2sqrt(13))/9#