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

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Answer 1 · 2016-12-16T09:09:06

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

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 $y in [(3-sqrt45)/2,(3+sqrt45)/2]$

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