How do you solve #-p^2+10p-7<=14#?

Question

1298 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-13T10:49:57

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

Let's rewrite the equation

#-p^2+10p-7<=14#, #hArr#, #p^2-10p+21>=0#

Let 's factorise the LHS

#(p-3)(p-7)>=0#

Let #f(p)=(p-3)(p-7)#

Now we can make the sign chart

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

#color(white)(aaaa)##p-3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##p-7##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

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