How do you solve #-2x^2+10<8x# using a sign chart?

1 Answer
Narad T.
Oct 29, 2016

The solution are #x<-5# and #x>1#

Explanation:

Let's rewrite the expression
#f(x)=2x^2+8x-10>0#
so factorising
#(2x-2)(x+5)=2(x-1)(x+5)>0#
So the values of x to be taken in consideration are #x=1# and #x=-5#

Let 's do the sign chart
#x##color(white)(aaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##1##color(white)(aaaa)##+oo#
#x-1##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#x+5##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#f(x)##color(white)(aaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

So the values of x are #x<-5# and #x>1#

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