How do you solve the inequality #x^2+ 3x>10#?

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Answer 1 · 2015-06-18T20:02:40

This inequality is equivalent to #(x+5)(x-2) > 0# which is satisfied for #x in (-oo, -5) uu (2, oo)#

First subtract #10# from both sides to get #x^2 + 3x - 10 > 0#

#f(x) = x^2 + 3x - 10# is a well behaved, continuous function, so the truth of the inequality will only change at the points where #f(x) = 0#.

#f(x) = x^2+3x - 10 = (x+5)(x-2)#

so #f(-5) = f(2) = 0#

Since we are looking for #f(x) > 0#, there are only three ranges we need to look at:

#(-oo, -5)# : #(x+5) < 0# and #(x-2) < 0#, so #f(x) > 0#

#(-5, 2)# : #(x+5) > 0#, #(x-2) < 0#, so #f(x) < 0#

#(2, oo)# : #(x+5) > 0# and #(x-2) > 0#, so #f(x) > 0#

Putting these cases together:

#f(x) > 0# for #x in (-oo, -5) uu (2, oo)#