How do you solve #x^2-3x-32>0#?

1 Answer
Jan 16, 2017

Answer:

#x > (3+sqrt(137))/2color(white)("XX")orcolor(white)("XX")x < (3-sqrt(137))/2#

Explanation:

If we use the quadratic formula to factor #color(red)1x^2color(blue)(-3)xcolor(green)(-32)#
we find zeros at
#color(white)("XXX")x=(-(color(blue)(-3))+-sqrt((color(blue)(-3))^2-4 * color(red)1 * (color(green)(-32))))/(2 * color(red)1)#
#color(white)("XXXX")=(3+-sqrt(137))/2#
giving the factors:
#color(white)("XXX")x^2-3x-32 =(x-(3+sqrt(137))/2) * (x - (3-sqrt(137))/2)#

#x^2-3x-32# will be greater than zero if and only if
#{: ("both terms are positive","or","both terms are negative"), (x>(3+sqrt(137))/2,,x < (3-sqrt(137))/2) :}#

If you like you could use a calculator to evaluate these irrational expressions:
#x > 7.352349955color(white)("XX")orcolor(white)("XX")x < -4.35234995#

We could verify that these results are reasonable by examining the graph of #x^2-3x-32# and noting the values of #x# for which the graph (#y#) is above the X-axis:
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