How do you solve #x^2+21>10x# using a sign chart?

1 Answer
Dec 30, 2016

Answer:

The answer is #x in ] -oo,3 [ uu ] 7, +oo[ #

Explanation:

Let's rearrange the equation

#x^2-10x+21>0#

Let #f(x)=x^2-10x+21#

The domain of #f(x)# is #D_f(x) = RR#

Let's factorise

#f(x)=(x-3)(x-7)#

Now, we can establish the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##3##color(white)(aaaaa)##7##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-3##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-7##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)>0# when # x in ] -oo,3 [ uu ] 7, +oo[ #