How do solve #x^2<10-3x# and write the answer as a inequality and interval notation?

1 Answer
Jan 31, 2017

Answer:

The answer is #-5 < x <2#
#x in ]-5, 2[#

Explanation:

Let's rewrite the inequality

#x^2+3x-10<0#

Let's factorise

#(x-2)(x+5)<0#

Let #f(x)=(x-2)(x+5)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##2##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

Therefore,

#f(x)<0# when #x in ]-5, 2[#

or #-5 < x <2#