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

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Answer 1 · 2016-12-22T05:55:08

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

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]}

How do you solve 15-2y-y^2<=015-2y-y^2<=0?