How do you solve #x^ { 2} + 40x - 96= 0#?

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Answer 1 · 2018-06-20T12:36:10

See a solution process below:

We can use the quadratic equation to solve this problem:

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)(40)# for #color(blue)(b)#

#color(green)(-96)# for #color(green)(c)# gives:

#x = (-color(blue)(40) +- sqrt(color(blue)(40)^2 - (4 * color(red)(1) * color(green)(-96))))/(2 * color(red)(1))#

#x = (-color(blue)(40) +- sqrt(1600 - (-384)))/2#

#x = (-color(blue)(40) +- sqrt(1600 + 384))/2#

#x = (-color(blue)(40) +- sqrt(1984))/2#

#x = (-color(blue)(40) +- sqrt(64 * 31))/2#

#x = (-color(blue)(40) +- sqrt(64)sqrt(31))/2#

#x = (-color(blue)(40) - sqrt(64)sqrt(31))/2#; #x = (-color(blue)(40) + sqrt(64)sqrt(31))/2#

#x = (-color(blue)(40) - 8sqrt(31))/2#; #x = (-color(blue)(40) + 8sqrt(31))/2#

#x = -color(blue)(40)/2 - (8sqrt(31))/2#; #x = -color(blue)(40)/2 + (8sqrt(31))/2#

#x = -20 - 4sqrt(31)#; #x = -20 + 4sqrt(31)#