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

1 Answer
Dec 22, 2016

Answer:

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]}