Question

How do you solve `15-2y-y^2<=0`?

Answer 1

The answer is `y in ] -oo,-5 ] uu [3, +oo[`

Explanation:

Let's factorise the equation

`15-2y-y^2=(5+y)(3-y)<=0`

Let `f(y)=(5+y)(3-y)`

Now, we can make a sign chart

`color(white)(aaaa)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `5+y` `color(white)(aaaa)` `-` `color(white)(aaa)` `0` `color(white)(aa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `3-y` `color(white)(aaaa)` `+` `color(white)(aaaaaa)` `+` `color(white)(a)` `0` `color(white)(aa)` `-`

`color(white)(aaaa)` `f(y)` `color(white)(aaaaa)` `-` `color(white)(aaaaaa)` `+` `color(white)(a)` ###color(white)(aaa)#`-`

Therefore,

`f(y)<=0`, when `y in ] -oo,-5 ] uu [3, +oo[`

graph{(5+x)(3-x) [-32.47, 32.48, -16.25, 16.24]}