How do you solve #2x^2+5x-12>0# using a sign chart?

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Answer 1 · 2016-12-01T06:39:17

The answer is #x in] -oo,-4 [ uu ] 3/2, oo[#

Let's start by solving the quadratic equation

#ax^2+bx+c=0#

Our equation is #f(x)=2x^2+5x-12#

Let's calculate the discriminant

#Delta=b^2-4ac=25-4*2(-12)=121#

#x=(-b+-sqrtDelta)/(2a)#

#x=(-5+-sqrt121)/4#

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

#x_1=-16/4=-4#

#x_2=6/4=3/2#

Let's do our sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##3/2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3/2##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaaa)##-##color(white)(aaaa)##+#

So, #f(x)>0#, when #x in] -oo,-4 [ uu ] 3/2, oo[ #