Question

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

Answer 1

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]`