How do you solve #y^2-3y-9<=0# using a sign chart?

1 Answer
Dec 16, 2016

Answer:

The answer is # y in [(3-sqrt45)/2,(3+sqrt45)/2] #

Explanation:

Let #f(y)=y^2-3y-9#

First we need the roots of the equation

#y^2-3y-9=0#

We calculate the discriminant

#Delta=b^2-4ac=(-3)^2-4*1+(-9)=45#

As Delta>0#. we have 2 real roots

#y=(3+-sqrtDelta)/2#

#y_2=(3+sqrt45)/2#

#y_1=(3-sqrt45)/2#

Now we can make the sign chart

#color(white)(aaaa)##y##color(white)(aaaa)##-oo##color(white)(aaaa)##y_1##color(white)(aaaa)##y_2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##y-y_1##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##y-y_2##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(y<=0)# when # y in [(3-sqrt45)/2,(3+sqrt45)/2] #