How do you solve and write the following in interval notation: #2x^2+5x-12 ≤ 0#?

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Answer 1 · 2016-06-21T11:43:55

#x in [-4;3/2]#

You must find the zeroes of the trynomial
#2x^2+5x-12#

using the formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case you have:

#a=1;b=5;c=-12#

so that

#x=(-5+-sqrt(25-4(2)(-12)))/(2(2))#

#x=(-5+-sqrt(25+96))/4#

#x=(-5+-sqrt(121))/4#

#x=(-5+-11)/4#

#x=-16/4=-4 and x=6/4=3/2#

The trynomial is therefore negative in the interval

#[-4;3/2]#